problem solving about projectile motion

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How to Solve a Projectile Motion Problem

Last Updated: April 6, 2024

This article was co-authored by Grace Imson, MA . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. This article has been viewed 77,051 times.

Projectile motion is often one of the most difficult topics to understand in physics classes. Most of the time, there is not a direct way to get the answer; you need to solve for a few other variables to get the answer you are looking for. This means in order to find the distance an object traveled, you might first have to find the time it took or the initial velocity first. Just follow these steps and you should be able to fly through projectile motion problems!

• (1) an object is thrown off a higher ground than what it will land on.
• (2) the object starts on the ground, soars through the air, and then lands on the ground some distance away from where it started.

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3.3: Projectile Motion

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Basic Equations and Parabolic Path

Projectile motion is a form of motion where an object moves in parabolic path; the path that the object follows is called its trajectory.

learning objectives

• Assess the effect of angle and velocity on the trajectory of the projectile; derive maximum height using displacement

Projectile Motion

Projectile motion is a form of motion where an object moves in a bilaterally symmetrical, parabolic path. The path that the object follows is called its trajectory. Projectile motion only occurs when there is one force applied at the beginning on the trajectory, after which the only interference is from gravity. In a previous atom we discussed what the various components of an object in projectile motion are. In this atom we will discuss the basic equations that go along with them in the special case in which the projectile initial positions are null (i.e. \(\mathrm{x_0=0}\) and \(\mathrm{y_0=0}\) ).

Initial Velocity

The initial velocity can be expressed as x components and y components:

\[\begin{align} \mathrm{u_x} & \mathrm{=u⋅ \cos θ} \\ \mathrm{u_y} & \mathrm{=u⋅ \sin θ} \end{align}\]

In this equation, \(\mathrm{u}\) stands for initial velocity magnitude and \(θ\) refers to projectile angle.

Time of Flight

The time of flight of a projectile motion is the time from when the object is projected to the time it reaches the surface. As we discussed previously, \(\mathrm{T}\) depends on the initial velocity magnitude and the angle of the projectile:

\[\begin{align} \mathrm{T} & \mathrm{=\dfrac{2⋅u_y}{g} } \\ \mathrm{T} & \mathrm{=\dfrac{2⋅u⋅\sin ⁡θ}{g}} \end{align}\]

Acceleration

In projectile motion, there is no acceleration in the horizontal direction. The acceleration, \(\mathrm{a}\), in the vertical direction is just due to gravity, also known as free fall:

\[\begin{align} \mathrm{a_x} & \mathrm{=0} \\ \mathrm{a_y} &\mathrm{=−g} \end{align}\]

The horizontal velocity remains constant, but the vertical velocity varies linearly, because the acceleration is constant. At any time, \(\mathrm{t}\), the velocity is:

\[\begin{align} \mathrm{u_x} & \mathrm{=u⋅ \cos θ} \\ \mathrm{u_y} & \mathrm{=u⋅ \sin θ−g⋅t} \end{align}\]

You can also use the Pythagorean Theorem to find velocity:

\[\mathrm{u=\sqrt{u_x^2+u_y^2}}\]

Displacement

At time, t, the displacement components are:

\[\begin{align} \mathrm{x} & \mathrm{=u⋅t⋅ \cos ⁡θ} \\ \mathrm{y} & \mathrm{=u⋅t⋅ \sin ⁡θ−\dfrac{1}{2}gt^2} \end{align}\]

The equation for the magnitude of the displacement is \(\mathrm{Δr=\sqrt{x^2+y^2}}\).

Parabolic Trajectory

We can use the displacement equations in the x and y direction to obtain an equation for the parabolic form of a projectile motion:

\[\mathrm{y=\tan θ⋅x−\dfrac{g}{2⋅u^2⋅ \cos ^2 θ}⋅x^2}\]

Maximum Height

The maximum height is reached when \(\mathrm{v_y=0}\). Using this we can rearrange the velocity equation to find the time it will take for the object to reach maximum height

\[\mathrm{t_h=\dfrac{u⋅\sin θ}{g}}\]

where \(\mathrm{t_h}\) stands for the time it takes to reach maximum height. From the displacement equation we can find the maximum height

\[\mathrm{h=\dfrac{u^2⋅ \sin ^2 ⁡θ}{2⋅g}}\]

The range of the motion is fixed by the condition \(\mathrm{y=0}\). Using this we can rearrange the parabolic motion equation to find the range of the motion:

\[\mathrm{R=\dfrac{u^2⋅ \sin ⁡2 θ}{g}.}\]

Range of Trajectory : The range of a trajectory is shown in this figure.

Projectiles at an Angle : This video gives a clear and simple explanation of how to solve a problem on Projectiles Launched at an Angle. I try to go step by step through this difficult problem to layout how to solve it in a super clear way. 2D kinematic problems take time to solve, take notes on the order of how I solved it. Best wishes. Tune into my other videos for more help. Peace.

Solving Problems

In projectile motion, an object moves in parabolic path; the path the object follows is called its trajectory.

• Identify which components are essential in determining projectile motion of an object

We have previously discussed projectile motion and its key components and basic equations. Using that information, we can solve many problems involving projectile motion. Before we do this, let’s review some of the key factors that will go into this problem-solving.

What is Projectile Motion?

Projectile motion is when an object moves in a bilaterally symmetrical, parabolic path. The path that the object follows is called its trajectory. Projectile motion only occurs when there is one force applied at the beginning, after which the only influence on the trajectory is that of gravity.

What are the Key Components of Projectile Motion?

The key components that we need to remember in order to solve projectile motion problems are:

• Initial launch angle, \(\mathrm{θ}\)
• Initial velocity, \(\mathrm{u}\)
• Time of flight, \(\mathrm{T}\)
• Acceleration, \(\mathrm{a}\)
• Horizontal velocity, \(\mathrm{v_x}\)
• Vertical velocity, \(\mathrm{v_y}\)
• Displacement, \(\mathrm{d}\)
• Maximum height, \(\mathrm{H}\)
• Range, \(\mathrm{R}\)

How To Solve Any Projectile Motion Problem (The Toolbox Method) : Introducing the “Toolbox” method of solving projectile motion problems! Here we use kinematic equations and modify with initial conditions to generate a “toolbox” of equations with which to solve a classic three-part projectile motion problem.

Now, let’s look at two examples of problems involving projectile motion.

Example \(\PageIndex{1}\):

Let’s say you are given an object that needs to clear two posts of equal height separated by a specific distance. Refer to for this example. The projectile is thrown at \(\mathrm{25 \sqrt{2}}\) m/s at an angle of 45°. If the object is to clear both posts, each with a height of 30m, find the minimum: (a) position of the launch on the ground in relation to the posts and (b) the separation between the posts. For simplicity’s sake, use a gravity constant of 10. Problems of any type in physics are much easier to solve if you list the things that you know (the “givens”).

Diagram for Example 1 : Use this figure as a reference to solve example 1. The problem is to make sure the object is able to clear both posts.

Solution: The first thing we need to do is figure out at what time tt the object reaches the specified height. Since the motion is in a parabolic shape, this will occur twice: once when traveling upward, and again when the object is traveling downward. For this we can use the equation of displacement in the vertical direction, \(\mathrm{y−y_0}\) :

\[\mathrm{y−y_0=(v_y⋅t)−(\dfrac{1}{2}⋅g⋅t^2)}\]

We substitute in the appropriate variables:

\[\mathrm{v_y=u⋅ \sin θ=25 \sqrt{2} \dfrac{m}{s}⋅ \sin 45^{\circ}=25 \dfrac{m}{s}}\]

\[\mathrm{30m=25⋅t−\dfrac{1}{2}⋅10⋅t^2}\]

We can use the quadratic equation to find that the roots of this equation are 2s and 3s. This means that the projectile will reach 30m after 2s, on its way up, and after 3s, on its way down.

An object is launched from the base of an incline, which is at an angle of 30°. If the launch angle is 60° from the horizontal and the launch speed is 10 m/s, what is the total flight time? The following information is given: \(\mathrm{u=10 \frac{m}{s}; θ=60°; g=10 \frac{m}{s^2}}\).

Diagram for Example 2 : When dealing with an object in projectile motion on an incline, we first need to use the given information to reorient the coordinate system in order to have the object launch and fall on the same surface.

Solution: In order to account for the incline angle, we have to reorient the coordinate system so that the points of projection and return are on the same level. The angle of projection with respect to the \(\mathrm{x}\) direction is \(\mathrm{θ−α}\), and the acceleration in the \(\mathrm{y}\) direction is \(\mathrm{g⋅ \cos ⁡α}\). We replace \(\mathrm{θ}\) with \(\mathrm{θ−α}\) and \(\mathrm{g}\) with \(\mathrm{g⋅ \cos ⁡α}\):

\[\begin{align} \mathrm{T \; } & \mathrm{=\dfrac{2⋅u⋅ \sin (θ)}{g} = \dfrac{2⋅u⋅\sin (θ−α)}{g⋅ \cos (α)}=\dfrac{2⋅10⋅ \sin (60−30)}{10⋅\cos (30)} =\dfrac{20⋅ \sin (30)}{10⋅ \cos (30)} } \\ \mathrm{T \;} &\mathrm{=\dfrac{2}{\sqrt{3}}s} \end{align}\]

Zero Launch Angle

An object launched horizontally at a height \(\mathrm{H}\) travels a range \(\mathrm{v_0\sqrt{\frac{2H}{g}}}\) during a time of flight \(\mathrm{T=\sqrt{\frac{2H}{g}}}\).

• Explain the relationship between the range and the time of flight

Projectile motion is a form of motion where an object moves in a parabolic path. The path followed by the object is called its trajectory. Projectile motion occurs when a force is applied at the beginning of the trajectory for the launch (after this the projectile is subject only to the gravity).

One of the key components of the projectile motion, and the trajectory it follows, is the initial launch angle . The angle at which the object is launched dictates the range, height, and time of flight the object will experience while in projectile motion. shows different paths for the same object being launched at the same initial velocity and different launch angles. As illustrated by the figure, the larger the initial launch angle and maximum height, the longer the flight time of the object.

Projectile Trajectories : The launch angle determines the range and maximum height that an object will experience after being launched.This image shows that path of the same object being launched at the same speed but different angles.

We have previously discussed the effects of different launch angles on range, height, and time of flight. However, what happens if there is no angle, and the object is just launched horizontally? It makes sense that the object should be launched at a certain height (\(\mathrm{H}\)), otherwise it wouldn’t travel very far before hitting the ground. Let’s examine how an object launched horizontally at a height \(\mathrm{H}\) travels. In our case is when \(\mathrm{α}\) is 0.

Projectile motion : Projectile moving following a parabola.Initial launch angle is αα, and the velocity is \(\mathrm{v_0}\).

Duration of Flight

There is no vertical component in the initial velocity (\(\mathrm{v_0}\)) because the object is launched horizontally. Since the object travels distance \(\mathrm{H}\) in the vertical direction before it hits the ground, we can use the kinematic equation for the vertical motion:

\[\mathrm{(y−y_0)=−H=0⋅T−\dfrac{1}{2}gT^2}\]

Here, \(\mathrm{T}\) is the duration of the flight before the object its the ground. Therefore:

\[\mathrm{T=\sqrt{\dfrac{2H}{g}}}\]

In the horizontal direction, the object travels at a constant speed \(\mathrm{v_0}\) during the flight. Therefore, the range \(\mathrm{R}\) (in the horizontal direction) is given as:

\[\mathrm{R=v_0⋅T=v_0\sqrt{\dfrac{2H}{g}}}\]

General Launch Angle

The initial launch angle (0-90 degrees) of an object in projectile motion dictates the range, height, and time of flight of that object.

• Choose the appropriate equation to find range, maximum height, and time of flight

Projectile motion is a form of motion where an object moves in a bilaterally symmetrical, parabolic path. The path that the object follows is called its trajectory. Projectile motion only occurs when there is one force applied at the beginning of the trajectory, after which the only interference is from gravity.

One of the key components of projectile motion and the trajectory that it follows is the initial launch angle. This angle can be anywhere from 0 to 90 degrees. The angle at which the object is launched dictates the range, height, and time of flight it will experience while in projectile motion. shows different paths for the same object launched at the same initial velocity at different launch angles. As you can see from the figure, the larger the initial launch angle, the closer the object comes to maximum height and the longer the flight time. The largest range will be experienced at a launch angle up to 45 degrees.

Launch Angle : The launch angle determines the range and maximum height that an object will experience after being launched. This image shows that path of the same object being launched at the same velocity but different angles.

The range, maximum height, and time of flight can be found if you know the initial launch angle and velocity, using the following equations:

\[\begin{align} \mathrm{R \;} & \mathrm{=\dfrac{v_i^2 \sin ^⁡2 θ_i}{g}} \\ \mathrm{h \;} & \mathrm{=\dfrac{v_i^2 \sin ^2 ⁡θ_i}{2g}} \\ \mathrm{T \;} & \mathrm{=\dfrac{2v_i \sin θ}{g}} \end{align}\]

Where R – Range, h – maximum height, T – time of flight, v i – initial velocity, θ i – initial launch angle, g – gravity.

Now that we understand how the launch angle plays a major role in many other components of the trajectory of an object in projectile motion, we can apply that knowledge to making an object land where we want it. If there is a certain distance, d, that you want your object to go and you know the initial velocity at which it will be launched, the initial launch angle required to get it that distance is called the angle of reach. It can be found using the following equation:

\[\mathrm{θ=\dfrac{1}{2} \sin ^{−1}(\dfrac{gd}{v^2})}\]

Key Points: Range, Symmetry, Maximum Height

Projectile motion is a form of motion where an object moves in parabolic path. The path that the object follows is called its trajectory.

• Construct a model of projectile motion by including time of flight, maximum height, and range

What is Projectile Motion ?

Projectile motion is a form of motion where an object moves in a bilaterally symmetrical, parabolic path. The path that the object follows is called its trajectory. Projectile motion only occurs when there is one force applied at the beginning on the trajectory, after which the only interference is from gravity. In this atom we are going to discuss what the various components of an object in projectile motion are, we will discuss the basic equations that go along with them in another atom, “Basic Equations and Parabolic Path”

Key Components of Projectile Motion:

Time of flight, t:.

The time of flight of a projectile motion is exactly what it sounds like. It is the time from when the object is projected to the time it reaches the surface. The time of flight depends on the initial velocity of the object and the angle of the projection, θθ. When the point of projection and point of return are on the same horizontal plane, the net vertical displacement of the object is zero.

All projectile motion happens in a bilaterally symmetrical path, as long as the point of projection and return occur along the same horizontal surface. Bilateral symmetry means that the motion is symmetrical in the vertical plane. If you were to draw a straight vertical line from the maximum height of the trajectory, it would mirror itself along this line.

Maximum Height, H:

The maximum height of a object in a projectile trajectory occurs when the vertical component of velocity, vyvy, equals zero. As the projectile moves upwards it goes against gravity, and therefore the velocity begins to decelerate. Eventually the vertical velocity will reach zero, and the projectile is accelerated downward under gravity immediately. Once the projectile reaches its maximum height, it begins to accelerate downward. This is also the point where you would draw a vertical line of symmetry.

Range of the Projectile, R:

The range of the projectile is the displacement in the horizontal direction. There is no acceleration in this direction since gravity only acts vertically. shows the line of range. Like time of flight and maximum height, the range of the projectile is a function of initial speed.

Range : The range of a projectile motion, as seen in this image, is independent of the forces of gravity.

• Objects that are projected from, and land on the same horizontal surface will have a vertically symmetrical path.
• The time it takes from an object to be projected and land is called the time of flight. This depends on the initial velocity of the projectile and the angle of projection.
• When the projectile reaches a vertical velocity of zero, this is the maximum height of the projectile and then gravity will take over and accelerate the object downward.
• The horizontal displacement of the projectile is called the range of the projectile, and depends on the initial velocity of the object.
• When solving problems involving projectile motion, we must remember all the key components of the motion and the basic equations that go along with them.
• Using that information, we can solve many different types of problems as long as we can analyze the information we are given and use the basic equations to figure it out.
• To clear two posts of equal height, and to figure out what the distance between these posts is, we need to remember that the trajectory is a parabolic shape and that there are two different times at which the object will reach the height of the posts.
• When dealing with an object in projectile motion on an incline, we first need to use the given information to reorientate the coordinate system in order to have the object launch and fall on the same surface.
• For the zero launch angle, there is no vertical component in the initial velocity.
• The duration of the flight before the object hits the ground is given as \(\mathrm{T=\sqrt{\frac{2H}{g}}}\).
• In the horizontal direction, the object travels at a constant speed v 0 during the flight. The range R (in the horizontal direction) is given as: \(\mathrm{R=v_0⋅T=v_0\sqrt{\dfrac{2H}{g}}}\).
• If the same object is launched at the same initial velocity, the height and time of flight will increase proportionally to the initial launch angle.
• An object launched into projectile motion will have an initial launch angle anywhere from 0 to 90 degrees.
• The range of an object, given the initial launch angle and initial velocity is found with: \(\mathrm{R=\dfrac{v_i^2 \sin ^⁡2 θ_i}{g}}\).
• The maximum height of an object, given the initial launch angle and initial velocity is found with: \(\mathrm{ｈ=\dfrac{v_i^2 \sin ^2 ⁡θ_i}{2g}}\).
• The time of flight of an object, given the initial launch angle and initial velocity is found with: \(\mathrm{T=\dfrac{2v_i \sin θ}{g}}\) .
• The angle of reach is the angle the object must be launched at in order to achieve a specific distance: \(\mathrm{θ=\dfrac{1}{2} \sin ^{−1}(\dfrac{gd}{v^2})}\).
• Objects that are projected from and land on the same horizontal surface will have a path symmetric about a vertical line through a point at the maximum height of the projectile.
• The time it takes from an object to be projected and land is called the time of flight. It depends on the initial velocity of the projectile and the angle of projection.
• The maximum height of the projectile is when the projectile reaches zero vertical velocity. From this point the vertical component of the velocity vector will point downwards.
• The horizontal displacement of the projectile is called the range of the projectile and depends on the initial velocity of the object.
• If an object is projected at the same initial speed, but two complementary angles of projection, the range of the projectile will be the same.
• trajectory : The path of a body as it travels through space.
• symmetrical : Exhibiting symmetry; having harmonious or proportionate arrangement of parts; having corresponding parts or relations.
• reorientate : to orientate anew; to cause to face a different direction
• gravity : Resultant force on Earth’s surface, of the attraction by the Earth’s masses, and the centrifugal pseudo-force caused by the Earth’s rotation.
• bilateral symmetry : the property of being symmetrical about a vertical plane

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4.3 Projectile Motion

Learning objectives.

By the end of this section, you will be able to:

• Use one-dimensional motion in perpendicular directions to analyze projectile motion.
• Calculate the range, time of flight, and maximum height of a projectile that is launched and impacts a flat, horizontal surface.
• Find the time of flight and impact velocity of a projectile that lands at a different height from that of launch.
• Calculate the trajectory of a projectile.

Projectile motion is the motion of an object thrown or projected into the air, subject only to acceleration as a result of gravity. The applications of projectile motion in physics and engineering are numerous. Some examples include meteors as they enter Earth’s atmosphere, fireworks, and the motion of any ball in sports. Such objects are called projectiles and their path is called a trajectory . The motion of falling objects as discussed in Motion Along a Straight Line is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, and our treatment neglects the effects of air resistance.

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. We discussed this fact in Displacement and Velocity Vectors , where we saw that vertical and horizontal motions are independent. The key to analyzing two-dimensional projectile motion is to break it into two motions: one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible because acceleration resulting from gravity is vertical; thus, there is no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x -axis and the vertical axis the y -axis. It is not required that we use this choice of axes; it is simply convenient in the case of gravitational acceleration. In other cases we may choose a different set of axes. Figure 4.11 illustrates the notation for displacement, where we define s → s → to be the total displacement, and x → x → and y → y → are its component vectors along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s , x , and y .

To describe projectile motion completely, we must include velocity and acceleration, as well as displacement. We must find their components along the x- and y -axes. Let’s assume all forces except gravity (such as air resistance and friction, for example) are negligible. Defining the positive direction to be upward, the components of acceleration are then very simple:

Because gravity is vertical, a x = 0 . a x = 0 . If a x = 0 , a x = 0 , this means the initial velocity in the x direction is equal to the final velocity in the x direction, or v x = v 0 x . v x = v 0 x . With these conditions on acceleration and velocity, we can write the kinematic Equation 4.11 through Equation 4.18 for motion in a uniform gravitational field, including the rest of the kinematic equations for a constant acceleration from Motion with Constant Acceleration . The kinematic equations for motion in a uniform gravitational field become kinematic equations with a y = − g , a x = 0 : a y = − g , a x = 0 :

Horizontal Motion

Vertical Motion

Using this set of equations, we can analyze projectile motion, keeping in mind some important points.

Problem-Solving Strategy

Projectile motion.

• Resolve the motion into horizontal and vertical components along the x - and y -axes. The magnitudes of the components of displacement s → s → along these axes are x and y. The magnitudes of the components of velocity v → v → are v x = v cos θ and v y = v sin θ , v x = v cos θ and v y = v sin θ , where v is the magnitude of the velocity and θ is its direction relative to the horizontal, as shown in Figure 4.12 .
• Treat the motion as two independent one-dimensional motions: one horizontal and the other vertical. Use the kinematic equations for horizontal and vertical motion presented earlier.
• Solve for the unknowns in the two separate motions: one horizontal and one vertical. Note that the only common variable between the motions is time t . The problem-solving procedures here are the same as those for one-dimensional kinematics and are illustrated in the following solved examples.
• Recombine quantities in the horizontal and vertical directions to find the total displacement s → s → and velocity v → . v → . Solve for the magnitude and direction of the displacement and velocity using s = x 2 + y 2 , Φ = tan −1 ( y / x ) , v = v x 2 + v y 2 , s = x 2 + y 2 , Φ = tan −1 ( y / x ) , v = v x 2 + v y 2 , where Φ is the direction of the displacement s → . s → .

Example 4.7

A fireworks projectile explodes high and away.

Because y 0 y 0 and v y v y are both zero, the equation simplifies to

Solving for y gives

Now we must find v 0 y , v 0 y , the component of the initial velocity in the y direction. It is given by v 0 y = v 0 sin θ 0 , v 0 y = v 0 sin θ 0 , where v 0 v 0 is the initial velocity of 70.0 m/s and θ 0 = 75 ° θ 0 = 75 ° is the initial angle. Thus,

Thus, we have

Note that because up is positive, the initial vertical velocity is positive, as is the maximum height, but the acceleration resulting from gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6-m/s initial vertical component of velocity reaches a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, so the initial velocity would have to be somewhat larger than that given to reach the same height.

(b) As in many physics problems, there is more than one way to solve for the time the projectile reaches its highest point. In this case, the easiest method is to use v y = v 0 y − g t . v y = v 0 y − g t . Because v y = 0 v y = 0 at the apex, this equation reduces to simply

This time is also reasonable for large fireworks. If you are able to see the launch of fireworks, notice that several seconds pass before the shell explodes. Another way of finding the time is by using y = y 0 + 1 2 ( v 0 y + v y ) t . y = y 0 + 1 2 ( v 0 y + v y ) t . This is left for you as an exercise to complete.

(c) Because air resistance is negligible, a x = 0 a x = 0 and the horizontal velocity is constant, as discussed earlier. The horizontal displacement is the horizontal velocity multiplied by time as given by x = x 0 + v x t , x = x 0 + v x t , where x 0 x 0 is equal to zero. Thus,

where v x v x is the x -component of the velocity, which is given by

Time t for both motions is the same, so x is

Horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. When the shell explodes, air resistance has a major effect, and many fragments land directly below.

(d) The horizontal and vertical components of the displacement were just calculated, so all that is needed here is to find the magnitude and direction of the displacement at the highest point:

Note that the angle for the displacement vector is less than the initial angle of launch. To see why this is, review Figure 4.11 , which shows the curvature of the trajectory toward the ground level.

When solving Example 4.7 (a), the expression we found for y is valid for any projectile motion when air resistance is negligible. Call the maximum height y = h . Then,

This equation defines the maximum height of a projectile above its launch position and it depends only on the vertical component of the initial velocity.

Check Your Understanding 4.3

A rock is thrown horizontally off a cliff 100.0 m 100.0 m high with a velocity of 15.0 m/s. (a) Define the origin of the coordinate system. (b) Which equation describes the horizontal motion? (c) Which equations describe the vertical motion? (d) What is the rock’s velocity at the point of impact?

Example 4.8

Calculating projectile motion: tennis player.

If we take the initial position y 0 y 0 to be zero, then the final position is y = 10 m. The initial vertical velocity is the vertical component of the initial velocity:

Substituting into Equation 4.22 for y gives us

Rearranging terms gives a quadratic equation in t :

Use of the quadratic formula yields t = 3.79 s and t = 0.54 s. Since the ball is at a height of 10 m at two times during its trajectory—once on the way up and once on the way down—we take the longer solution for the time it takes the ball to reach the spectator:

The time for projectile motion is determined completely by the vertical motion. Thus, any projectile that has an initial vertical velocity of 21.2 m/s and lands 10.0 m above its starting altitude spends 3.79 s in the air.

(b) We can find the final horizontal and vertical velocities v x v x and v y v y with the use of the result from (a). Then, we can combine them to find the magnitude of the total velocity vector v → v → and the angle θ θ it makes with the horizontal. Since v x v x is constant, we can solve for it at any horizontal location. We choose the starting point because we know both the initial velocity and the initial angle. Therefore,

The final vertical velocity is given by Equation 4.21 :

Since v 0 y v 0 y was found in part (a) to be 21.2 m/s, we have

The magnitude of the final velocity v → v → is

The direction θ v θ v is found using the inverse tangent:

Significance

Time of flight, trajectory, and range.

Of interest are the time of flight, trajectory, and range for a projectile launched on a flat horizontal surface and impacting on the same surface. In this case, kinematic equations give useful expressions for these quantities, which are derived in the following sections.

Time of flight

We can solve for the time of flight of a projectile that is both launched and impacts on a flat horizontal surface by performing some manipulations of the kinematic equations. We note the position and displacement in y must be zero at launch and at impact on an even surface. Thus, we set the displacement in y equal to zero and find

Factoring, we have

Solving for t gives us

This is the time of flight for a projectile both launched and impacting on a flat horizontal surface. Equation 4.24 does not apply when the projectile lands at a different elevation than it was launched, as we saw in Example 4.8 of the tennis player hitting the ball into the stands. The other solution, t = 0, corresponds to the time at launch. The time of flight is linearly proportional to the initial velocity in the y direction and inversely proportional to g . Thus, on the Moon, where gravity is one-sixth that of Earth, a projectile launched with the same velocity as on Earth would be airborne six times as long.

The trajectory of a projectile can be found by eliminating the time variable t from the kinematic equations for arbitrary t and solving for y ( x ). We take x 0 = y 0 = 0 x 0 = y 0 = 0 so the projectile is launched from the origin. The kinematic equation for x gives

Substituting the expression for t into the equation for the position y = ( v 0 sin θ 0 ) t − 1 2 g t 2 y = ( v 0 sin θ 0 ) t − 1 2 g t 2 gives

Rearranging terms, we have

This trajectory equation is of the form y = a x + b x 2 , y = a x + b x 2 , which is an equation of a parabola with coefficients

From the trajectory equation we can also find the range , or the horizontal distance traveled by the projectile. Factoring Equation 4.25 , we have

The position y is zero for both the launch point and the impact point, since we are again considering only a flat horizontal surface. Setting y = 0 in this equation gives solutions x = 0, corresponding to the launch point, and

corresponding to the impact point. Using the trigonometric identity 2 sin θ cos θ = sin 2 θ 2 sin θ cos θ = sin 2 θ and setting x = R for range, we find

Note particularly that Equation 4.26 is valid only for launch and impact on a horizontal surface. We see the range is directly proportional to the square of the initial speed v 0 v 0 and sin 2 θ 0 sin 2 θ 0 , and it is inversely proportional to the acceleration of gravity. Thus, on the Moon, the range would be six times greater than on Earth for the same initial velocity. Furthermore, we see from the factor sin 2 θ 0 sin 2 θ 0 that the range is maximum at 45 ° . 45 ° . These results are shown in Figure 4.15 . In (a) we see that the greater the initial velocity, the greater the range. In (b), we see that the range is maximum at 45 ° . 45 ° . This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is somewhat smaller. It is interesting that the same range is found for two initial launch angles that sum to 90 ° . 90 ° . The projectile launched with the smaller angle has a lower apex than the higher angle, but they both have the same range.

Example 4.9

Comparing golf shots.

(a) What is the initial speed of the ball at the second hole?

(b) What is the initial speed of the ball at the fourth hole?

(c) Write the trajectory equation for both cases.

(d) Graph the trajectories.

(b) R = v 0 2 sin 2 θ 0 g ⇒ v 0 = R g sin 2 θ 0 = 90.0 m ( 9.8 m / s 2 ) sin ( 2 ( 70 ° ) ) = 37.0 m / s R = v 0 2 sin 2 θ 0 g ⇒ v 0 = R g sin 2 θ 0 = 90.0 m ( 9.8 m / s 2 ) sin ( 2 ( 70 ° ) ) = 37.0 m / s

(c) y = x [ tan θ 0 − g 2 ( v 0 cos θ 0 ) 2 x ] Second hole: y = x [ tan 30 ° − 9.8 m / s 2 2 [ ( 31.9 m / s)( cos 30 ° ) ] 2 x ] = 0.58 x − 0.0064 x 2 Fourth hole: y = x [ tan 70 ° − 9.8 m / s 2 2 [ ( 37.0 m / s)( cos 70 ° ) ] 2 x ] = 2.75 x − 0.0306 x 2 y = x [ tan θ 0 − g 2 ( v 0 cos θ 0 ) 2 x ] Second hole: y = x [ tan 30 ° − 9.8 m / s 2 2 [ ( 31.9 m / s)( cos 30 ° ) ] 2 x ] = 0.58 x − 0.0064 x 2 Fourth hole: y = x [ tan 70 ° − 9.8 m / s 2 2 [ ( 37.0 m / s)( cos 70 ° ) ] 2 x ] = 2.75 x − 0.0306 x 2

(d) Using a graphing utility, we can compare the two trajectories, which are shown in Figure 4.16 .

Check Your Understanding 4.4

If the two golf shots in Example 4.9 were launched at the same speed, which shot would have the greatest range?

When we speak of the range of a projectile on level ground, we assume R is very small compared with the circumference of Earth. If, however, the range is large, Earth curves away below the projectile and the acceleration resulting from gravity changes direction along the path. The range is larger than predicted by the range equation given earlier because the projectile has farther to fall than it would on level ground, as shown in Figure 4.17 , which is based on a drawing in Newton’s Principia. If the initial speed is great enough, the projectile goes into orbit. Earth’s surface drops 5 m every 8000 m. In 1 s an object falls 5 m without air resistance. Thus, if an object is given a horizontal velocity of 8000 m/s (or 18,000 mi/hr) near Earth’s surface, it will go into orbit around the planet because the surface continuously falls away from the object. This is roughly the speed of the Space Shuttle in a low Earth orbit when it was operational, or any satellite in a low Earth orbit. These and other aspects of orbital motion, such as Earth’s rotation, are covered in greater depth in Gravitation .

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At PhET Explorations: Projectile Motion , learn about projectile motion in terms of the launch angle and initial velocity.

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• Authors: William Moebs, Samuel J. Ling, Jeff Sanny
• Publisher/website: OpenStax
• Book title: University Physics Volume 1
• Publication date: Sep 19, 2016
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• Section URL: https://openstax.org/books/university-physics-volume-1/pages/4-3-projectile-motion

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Two-Dimensional Kinematics

Projectile motion, learning objectives.

By the end of this section, you will be able to:

• Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory.
• Determine the location and velocity of a projectile at different points in its trajectory.
• Apply the principle of independence of motion to solve projectile motion problems.

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile , and its path is called its trajectory . The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional Kinematics, is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance  is negligible .

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction , where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x -axis and the vertical axis the y -axis. Figure 1 illustrates the notation for displacement, where s is defined to be the total displacement and x and y are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s , x , and y . (Note that in the last section we used the notation A to represent a vector with components A x and A y . If we continued this format, we would call displacement s with components s x and s y . However, to simplify the notation, we will simply represent the component vectors as x and y .)

Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x – and y -axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: a y  = – g  = –9.80 m/s 2 . (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, a x =0. Both accelerations are constant, so the kinematic equations can be used.

Review of Kinematic Equations (constant a )

Figure 1. The total displacement s of a soccer ball at a point along its path. The vector s has components x and y along the horizontal and vertical axes. Its magnitude is s, and it makes an angle θ with the horizontal.

Given these assumptions, the following steps are then used to analyze projectile motion:

Step 1.  Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so A x = A cos θ  and  A y  = A sin θ  are used. The magnitude of the components of displacement s along these axes are x and y. The magnitudes of the components of the velocity v are V x = V cos θ  and V y = v sin θ  where v is the magnitude of the velocity and θ is its direction, as shown in 2. Initial values are denoted with a subscript 0, as usual.

Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time t . The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below.

Step 4.  Recombine the two motions to find the total displacement s and velocity v. Because the x – and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing [latex]A=\sqrt{{{A}_{x}}^{2}+{{A}_{y}}^{2}}\\[/latex] and θ  = tan −1  ( A y / A x ) in the following form, where θ is the direction of the displacement s and θ v is the direction of the velocity v :

Total displacement and velocity

Figure 2. (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because ax=0 and vx is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x – and y -motions are recombined to give the total velocity at any given point on the trajectory.

Example 1. A Fireworks Projectile Explodes High and Away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0º above the horizontal, as illustrated in Figure 3. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which a x  = 0 and a y  = – g . We can then define x 0 and y 0 to be zero and solve for the desired quantities.

Solution for (a)

By “height” we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, called the apex, is reached when v y =0. Since we know the initial and final velocities as well as the initial position, we use the following equation to find y :

Figure 3. The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a height of 233 m and 125 m away horizontally.

Because y 0 and v y are both zero, the equation simplifies to

Solving for y gives

Now we must find v 0 y , the component of the initial velocity in the y -direction. It is given by v 0y = v 0 sin  θ , where v 0 y is the initial velocity of 70.0 m/s, and θ 0  = 75.0º is the initial angle. Thus,

v Oy = v 0 sin  θ 0 = (70.0 m/s)(sin 75º) = 67.6 m/s

[latex]y=\frac{\left(67.6\text{ m/s}\right)^{2}}{2\left(9.80\text{ m/s}^{2}\right)}\\[/latex] ,

Discussion for (a)

Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height.

Solution for (b)

As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use [latex]y={y}_{0}+\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t\\[/latex]. Because y 0 is zero, this equation reduces to simply

[latex]y=\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t\\[/latex].

Note that the final vertical velocity, v y , at the highest point is zero. Thus,

[latex]\begin{array}{lll}t& =& \frac{2y}{\left({v}_{0y}+{v}_{y}\right)}=\frac{2\left(\text{233 m}\right)}{\left(\text{67.6 m/s}\right)}\\ & =& 6.90\text{ s}\end{array}\\[/latex].

Discussion for (b)

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using [latex]y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\text{gt}}^{2}\\[/latex], and solving the quadratic equation for t .)

Solution for (c)

Because air resistance is negligible, a x =0 and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by x  =  x 0  +  v x t , where x 0 is equal to zero:

x  =  v x t ,

where v x is the x -component of the velocity, which is given by v x = v 0 cos θ 0  Now,

v x  = v 0  cos  θ 0 = (70.0 m/s)(cos 75º) = 18.1 m/s

The time t for both motions is the same, and so x is

x = (18.1 m/s)(6.90 s) = 125 m.

Discussion for (c)

The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below.

In solving part (a) of the preceding example, the expression we found for y is valid for any projectile motion where air resistance is negligible. Call the maximum height y = h ; then,

[latex]h=\frac{{{v}_{0y}}^{2}}{2g}\\[/latex].

This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity.

Defining a Coordinate System

Example 2. calculating projectile motion: hot rock projectile.

Kilauea in Hawaii is the world’s most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an angle 35.0º above the horizontal, as shown in Figure 4. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock’s velocity at impact?

Figure 4. The trajectory of a rock ejected from the Kilauea volcano.

Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for t first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain v and θ v at the final time t determined in the first part of the example.

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

[latex]y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\text{gt}}^{2}\\[/latex].

If we take the initial position y 0 to be zero, then the final position is y  = −20.0 m. Now the initial vertical velocity is the vertical component of the initial velocity, found from  v Oy = v 0 sin  θ 0  = (25.0 m/s)(sin 35.0º) = 14.3 m/s. Substituting known values yields

Rearranging terms gives a quadratic equation in t :

This expression is a quadratic equation of the form at 2 + bt + c = 0 , where the constants are a  = 4.90 , b  = –14.3 , and c  = –20.0. Its solutions are given by the quadratic formula:

[latex]t=\frac{-bpm \sqrt{{b}^{2}-4\text{ac}}}{\text{2}\text{a}}\\[/latex]

This equation yields two solutions: t  = 3.96 and t  = –1.03. (It is left as an exercise for the reader to verify these solutions.) The time is t = 3.96 s or -1.03 s. The negative value of time implies an event before the start of motion, and so we discard it. Thus,

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

From the information now in hand, we can find the final horizontal and vertical velocities v x and v y and combine them to find the total velocity v and the angle θ 0 it makes with the horizontal. Of course, v x is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore:

  v x  = v 0  cos  θ 0  = (25.0 m/s)(cos 35º) = 20.5 m/s

The final vertical velocity is given by the following equation:

[latex]{v}_{y}={v}_{0y}\text{gt}\\[/latex],

where v 0y was found in part (a) to be 14.3 m/s. Thus,

To find the magnitude of the final velocity v we combine its perpendicular components, using the following equation:

[latex]v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}=\sqrt{({20.5}\text{ m/s})^{2}+{({-24.5}\text{ m/s})^{2}}}\\[/latex]

which gives

The direction θ v is found from the equation:

The negative angle means that the velocity is 50.1º below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure 4.)

Figure 5. Trajectories of projectiles on level ground. (a) The greater the initial speed v0, the greater the range for a given initial angle. (b) The effect of initial angle θ 0  on the range of a projectile with a given initial speed. Note that the range is the same for 15º and 75º, although the maximum heights of those paths are different.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed v 0 , the greater the range, as shown in Figure 5(a). The initial angle θ 0 also has a dramatic effect on the range, as illustrated in Figure 5(b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with θ 0  = 45º. This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately 38º. Interestingly, for every initial angle except 45º, there are two angles that give the same range—the sum of those angles is 90º. The range also depends on the value of the acceleration of gravity g . The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range R of a projectile on level ground for which air resistance is negligible is given by

[latex]R=\frac{{{v}_{0}}^{2}\sin{2\theta }_{0}}{g}\\[/latex],

where v 0 is the initial speed and θ 0 is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described. When we speak of the range of a projectile on level ground, we assume that R is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure 6.) If the initial speed is great enough, the projectile goes into orbit. This is called escape velocity. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text. Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of Velocities , we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic.

Figure 6. Projectile to satellite. In each case shown here, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than it would be on level ground because the Earth curves away underneath its path. With a large enough initial speed, orbit is achieved.

PhET Explorations: Projectile Motion

Click to run the simulation.

Section Summary

• Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity.
• To solve projectile motion problems, perform the following steps:

1. Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position s are given by the quantities x  and y , and the components of the velocity v are given by v x =   v cos θ  and v y = v sin θ , where v  is the magnitude of the velocity and  θ  is its direction.

2. Analyze the motion of the projectile in the horizontal direction using the following equations:

Horizontal Motion ( a x = 0)

3. Analyze the motion of the projectile in the vertical direction using the following equations:

Vertical Motion (assuming positive is up a y = -g = -9.8 m/s 2 )

4. Recombine the horizontal and vertical components of location and/or velocity using the following equations:

• The maximum height  h of a projectile launched with initial vertical velocity v 0y is given by [latex]h=\frac{{{v}_{0y}}^{2}}{2g}\\[/latex].
• The maximum horizontal distance traveled by a projectile is called the range . The range R of a projectile on level ground launched at an angle  θ 0   above the horizontal with initial speed v 0 is given by [latex]R=\frac{{{{v}_{0}}^{2}}\text{\sin}{2\theta }_{0}}{g}\\[/latex].

Conceptual Questions

1. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0º nor 90º): (a) Is the velocity ever zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity ever be the same as the initial velocity at a time other than at t = 0? (d) Can the speed ever be the same as the initial speed at a time other than at  t = 0?

2. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0º nor 90º): (a) Is the acceleration ever zero? (b) Is the acceleration ever in the same direction as a component of velocity? (c) Is the acceleration ever opposite in direction to a component of velocity?

3. For a fixed initial speed, the range of a projectile is determined by the angle at which it is fired. For all but the maximum, there are two angles that give the same range. Considering factors that might affect the ability of an archer to hit a target, such as wind, explain why the smaller angle (closer to the horizontal) is preferable. When would it be necessary for the archer to use the larger angle? Why does the punter in a football game use the higher trajectory?

4. During a lecture demonstration, a professor places two coins on the edge of a table. She then flicks one of the coins horizontally off the table, simultaneously nudging the other over the edge. Describe the subsequent motion of the two coins, in particular discussing whether they hit the floor at the same time.

Problems & Exercises

1. A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0º above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

2. A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c)What maximum height is attained by the ball?

3. A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

4. (a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32º ramp at a speed of 40.0 m/s (144 km/h). How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 20.0 m long? (b) Discuss what your answer implies about the margin of error in this act—that is, consider how much greater the range is than the horizontal distance he must travel to miss the end of the last bus. (Neglect air resistance.)

5. An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

6. A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand. (a) At what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used? (b) What other angle gives the same range, and why would it not be used? (c) How long did this pass take?

7. Verify the ranges for the projectiles in Figure 5 (a) for θ = 45º and the given initial velocities.

8. Verify the ranges shown for the projectiles in Figure 5(b) for an initial velocity of 50 m/s at the given initial angles.

9. The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.) (c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is 6.37 × 10 3 . How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?

10. An arrow is shot from a height of 1.5 m toward a cliff of height  H . It is shot with a velocity of 30 m/s at an angle of 60º above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow’s impact speed just before hitting the cliff?

11. In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

12. The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.

13. Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle  θ  below the horizontal. The service line is 11.9 m from the net, which is 0.91 m high. What is the angle  θ such that the ball just crosses the net? Will the ball land in the service box, whose out line is 6.40 m from the net?

14. A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. (a) If the ball is thrown at an angle of 25º relative to the ground and is caught at the same height as it is released, what is its initial speed relative to the ground? (b) How long does it take to get to the receiver? (c) What is its maximum height above its point of release?

15. Gun sights are adjusted to aim high to compensate for the effect of gravity, effectively making the gun accurate only for a specific range. (a) If a gun is sighted to hit targets that are at the same height as the gun and 100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 m away? The muzzle velocity of the bullet is 275 m/s. (b) Discuss qualitatively how a larger muzzle velocity would affect this problem and what would be the effect of air resistance.

16. An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her talons wiggles loose and falls into the lake 5.00 m below. Calculate the velocity of the fish relative to the water when it hits the water.

17. An owl is carrying a mouse to the chicks in its nest. Its position at that time is 4.00 m west and 12.0 m above the center of the 30.0 cm diameter nest. The owl is flying east at 3.50 m/s at an angle 30.0º below the horizontal when it accidentally drops the mouse. Is the owl lucky enough to have the mouse hit the nest? To answer this question, calculate the horizontal position of the mouse when it has fallen 12.0 m.

18. Suppose a soccer player kicks the ball from a distance 30 m toward the goal. Find the initial speed of the ball if it just passes over the goal, 2.4 m above the ground, given the initial direction to be 40º above the horizontal.

19. Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s.

20. The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 7.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved in solving projectile motion problems.

21. In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77 m. What was the initial speed of the shot if he released it at a height of 2.10 m and threw it at an angle of 38.0º above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at 45º  when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus, 38º  will give a longer range than 45º  in the shot put.)

22. A basketball player is running at 5.00 m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity. (a) What vertical velocity does he need to rise 0.750 m above the floor? (b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket?

23. A football player punts the ball at a 45º angle. Without an effect from the wind, the ball would travel 60.0 m horizontally. (a) What is the initial speed of the ball? (b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.50 m/s. What distance does the ball travel horizontally?

24. Prove that the trajectory of a projectile is parabolic, having the form [latex]y=\text{ax}+{\text{bx}}^{2}\\[/latex]. To obtain this expression, solve the equation [latex]x={v}_{0x}t\\[/latex] for t  and substitute it into the expression for [latex]y={v}_{0y}t-\left(1/2\right){\text{gt}}^{2}\\[/latex]. (These equations describe the x  and y positions of a projectile that starts at the origin.) You should obtain an equation of the form [latex]y=\text{ax}+{\text{bx}}^{2}\\[/latex] where a and b are constants.

25. Derive [latex]R=\frac{{{v}_{0}}^{2}\text{\sin}{2\theta }_{0}}{g}\\[/latex] for the range of a projectile on level ground by finding the time t at which y becomes zero and substituting this value of t  into the expression for x – x 0 , noting that R = x – x 0 .

26.  Unreasonable Results (a) Find the maximum range of a super cannon that has a muzzle velocity of 4.0 km/s. (b) What is unreasonable about the range you found? (c) Is the premise unreasonable or is the available equation inapplicable? Explain your answer. (d) If such a muzzle velocity could be obtained, discuss the effects of air resistance, thinning air with altitude, and the curvature of the Earth on the range of the super cannon.

27. Construct Your Own Problem Consider a ball tossed over a fence. Construct a problem in which you calculate the ball’s needed initial velocity to just clear the fence. Among the things to determine are; the height of the fence, the distance to the fence from the point of release of the ball, and the height at which the ball is released. You should also consider whether it is possible to choose the initial speed for the ball and just calculate the angle at which it is thrown. Also examine the possibility of multiple solutions given the distances and heights you have chosen.

Selected Solutions to Problems & Exercises

1.  x = 1.30 m × 10 2 , y = 30.9 m

3. (a) 3.50 s (b) 28.6 m/s (c) 34.3 m/s (d) 44.7 m/s, 50.2º below horizontal

5. (a) 18.4º (b) The arrow will go over the branch.

7. [latex]R=\frac{{{{v}_{0}}}^{}}{\sin{2\theta }_{0}g}\\[/latex]

For θ = 45º, [latex]R=\frac{{{{v}_{0}}}^{2}}{g}\\[/latex]

R = 91.9 m for v 0  = 30 m/s; R = 163 m for v 0 ; R = 255 m for v 0 = 50 m/s

9. (a) 560 m/s (b) 800 × 10 3 m  (c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b).

11. 1.50 m, assuming launch angle of 45º

13.  θ  =6.1º. Yes, the ball lands at 5.3 m from the net

15. (a) −0.486 m (b) The larger the muzzle velocity, the smaller the deviation in the vertical direction, because the time of flight would be smaller. Air resistance would have the effect of decreasing the time of flight, therefore increasing the vertical deviation.

17. 4.23 m. No, the owl is not lucky; he misses the nest.

19. No, the maximum range (neglecting air resistance) is about 92 m.

21. 15.0 m/s

23. (a) 24.2 m/s (b) The ball travels a total of 57.4 m with the brief gust of wind.

25. [latex]y-{y}_{0}=0={v}_{0y}t-\frac{1}{2}{gt}^{2}=\left({v}_{0}\sin\theta\right)t-\frac{1}{2}{gt}^{2}\\[/latex] ,

so that [latex]t=\frac{2\left({v}_{0}\sin\theta \right)}{g}\\[/latex]

[latex]x-{x}_{0}={v}_{0x}t=\left({v}_{0}\cos\theta \right)t=R\\[/latex], and substituting for t  gives:

[latex]R={v}_{0}\cos\theta \left(\frac{{2v}_{0}\sin\theta}{g}\right)=\frac{{{2v}_{0}}^{2}\sin\theta \cos\theta }{g}\\[/latex]

since [latex]2\sin\theta \cos\theta =\sin 2\theta\\[/latex], the range is:

[latex]R=\frac{{{v}_{0}}^{2}\sin 2\theta }{g}\\[/latex].

• College Physics. Authored by : OpenStax College. Located at : http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a/College_Physics . License : CC BY: Attribution . License Terms : Located at License
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Projectile Problems with Solutions and Explanations

Projectile problems are presented along with detailed solutions . These problems may be better understood when projectile equations are first reviewed. An interactive html 5 applet may be used to better understand the projectile equations.

Problem 1: An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal. a) What is the maximum height reached by the object? b) What is the total flight time (between launch and touching the ground) of the object? c) What is the horizontal range (maximum x above ground) of the object? d) What is the magnitude of the velocity of the object just before it hits the ground? Solution to Problem 1

Problem 5: A ball kicked from ground level at an initial velocity of 60 m/s and an angle θ with ground reaches a horizontal distance of 200 meters. a) What is the size of angle θ? b) What is time of flight of the ball? Solution to Problem 5

Problem 6: A ball of 600 grams is kicked at an angle of 35° with the ground with an initial velocity V 0 . a) What is the initial velocity V 0 of the ball if its kinetic energy is 22 Joules when its height is maximum? b) What is the maximum height reached by the ball Solution to Problem 6

Problem 7: A projectile starting from ground hits a target on the ground located at a distance of 1000 meters after 40 seconds. a) What is the size of the angle θ? b) At what initial velocity was the projectile launched? Solution to Problem 7

Problem 8: The trajectory of a projectile launched from ground is given by the equation y = -0.025 x 2 + 0.5 x, where x and y are the coordinate of the projectile on a rectangular system of axes. a) Find the initial velocity and the angle at which the projectile is launched. Solution to Problem 8

Problem 9: Two balls A and B of masses 100 grams and 300 grams respectively are pushed horizontally from a table of height 3 meters. Ball has is pushed so that its initial velocity is 10 m/s and ball B is pushed so that its initial velocity is 15 m/s. a) Find the time it takes each ball to hit the ground. b) What is the difference in the distance between the points of impact of the two balls on the ground? Solution to Problem 9

More References and links

• Velocity and Speed: Tutorials with Examples
• Velocity and Speed: Problems with Solutions
• Acceleration: Tutorials with Examples
• Uniform Acceleration Motion: Problems with Solutions
• Uniform Acceleration Motion: Equations with Explanations

• Exam Center
• Ticket Center
• Flash Cards
• Straight Line Motion

Grade 12: Physics Worksheet on Projectile Motion

Looking to master projectile motion in your physics class? Check out our comprehensive worksheet with detailed solutions to help you understand the concepts and excel in your studies!

There is also a pdf  for worksheets with answers.

A summary of projectile motion:

Any motion having the following conditions is called a projectile motion.     (i) Follows a parabolic path (trajectory).     (ii) Moves under the influence of gravity. 

The projectile motion formulas applied to solve two-dimensional projectile motion problems are as follows \begin{gather*} x=(v_0\cos\theta)t+x_0\\\\y=-\frac 12 gt^2+(v_0\sin\theta)t+y_0\\\\ v_y=v_0\sin\theta-gt\\\\v_y^2-(v_0\sin\theta)^2=-2g(y-y_0)\end{gather*} 

Projectile motion problems and answers

Problem (1): A person kicks a ball with an initial velocity of 15 m/s at an angle of 37° above the horizontal (neglecting the air resistance). Find  (a) the total time the ball is in the air.  (b) the horizontal distance traveled by the ball

Solution : The initial step in answering any projectile motion questions is to establish a coordinate system and sketch the path of the projectile, including its initial and final positions and velocities. 

By doing so, you will be able to solve the relevant projectile equations easily. 

Hence, we choose the origin of the coordinate system to be at the throwing point, $x_0=0, y_0=0$. 

(a) Here, we are seeking the time it takes for the projectile to travel from the point of release to the ground.  

In reality, projectiles have two independent motions: one in the horizontal direction with uniform motion at a constant velocity (i.e., $a_x=0$), and the other in the vertical direction under the effect of gravity (with $a_y=-g$).  

The kinematic equations that describe the horizontal and vertical distances are as follows:  \begin{gather*} x=x_0+(\underbrace{v_0\cos \theta}_{v_{0x}})t \\ y=-\frac 12 gt^2+(\underbrace{v_0\sin \theta}_{v_{0y}})t+y_0\end{gather*} By substituting the coordinates of the initial and final points into the vertical equation, we can find the total time the ball is in the air.

Setting $y=0$ in the second equation (because the projectile lands at the same level as the throwing point.), we have: \begin{align*} y&=-\frac 12 gt^2+(v_0\sin \theta)t+y_0\\\\ 0&=-\frac 12 (9.8)t^2+(15)\sin 37^\circ\,t+0 \end{align*} By rearranging the above expression, we can obtain two solutions for $t$: \begin{gather*} t_1=0 \\\\ t_2=\frac{2\times 15\sin37^\circ}{9.8}=1.84\,{\rm s}\end{gather*} The first time represents the starting moment, and the second represents the total time the ball was in the air. 

(b) As previously mentioned, projectile motion consists of two independent motions with different positions, velocities, and accelerations, each described by distinct kinematic equations. 

The time it takes for the projectile to reach a specific point horizontally is equal to the time it takes to fall vertically to that point. Therefore, time is the common factor in both the horizontal and vertical motions of a projectile.

In this particular problem, the time calculated in part (a) can be used in the horizontal kinematic equation to determine the distance traveled. 

Substituting the obtained time into the equation, we find that the projectile travels a distance of 22.08 meters. \begin{align*} x&=x_0+(v_0\cos \theta)t \\ &=0+(15)\cos 37^\circ\,(1.84) \\ &=22.08\quad {\rm m}\end{align*}

Be sure to check the following questions.

If you are getting ready for the AP Physics exam, these problems on the kinematics equation (or these AP Physics Kinematics Problems ) are probably also useful to you. 

Problem (2): A ball is thrown into the air at an angle of 60° above the horizontal with an initial velocity of 40 m/s from a  50-m-high building. Find  (a) The time to reach the top of its path. (b) The maximum height the ball reached from the base of the building.

Solution : In the previous question, we found that the motion of a projectile consists of two distinct vertical and horizontal movements.

We learned about how to find distances in both directions using relevant kinematic equations. 

There is also another set of kinematic equations that discuss the velocities in vertical and horizontal directions as follows \begin{gather*} v_x=v_{0x}=v_0\cos\theta \\ v_y=v_{0y}-gt=v_0\sin\theta-gt\end{gather*} As you can see, the horizontal component of the velocity, $v_x$, is constant throughout the motion, but the vertical component varies with time. 

As an important note, keep in mind that in the problems about projectile motions, at the highest point of the trajectory, the vertical component of the velocity is always zero, i.e., $v_y=0$. 

To solve the first part of the problem, specify two points as the initial and final points, then solve the relevant kinematic equations between those points.

Here, setting $v_y=0$ in the second equation and solving for the unknown time $t$, we have \begin{align*} v_y&=v_{0y}-gt \\\\ &=v_0\sin\theta-gt \\\\ 0&=(40)(\sin 60^\circ)-(9.8)(t) \\\\ \Rightarrow t&=\boxed{3.53\, {\rm s}}\end{align*} Thus, the time taken to reach the maximum height of the trajectory (path) is 3.53 seconds from the moment of the launch. We call this maximum time $t_{max}$. 

(b) Let the origin be at the throwing point, so $x_0=y_0=0$ in the kinematic equations. In this part, the vertical distance traveled to the maximum point is requested. 

By substituting $t_{max}$ into the vertical distance projectile equation, we can find the maximum height as below \begin{align*} y-y_0&=-\frac 12 gt^2 +(v_0 \sin \theta)t\\ \\ y_{max}&=-\frac 12 (9.8)(3.53)^2+(40 \sin 60^\circ)(3.53)\\\\ &=61.22\quad {\rm m}\end{align*} Therefore, the maximum height at which the ball is reached from the base of the building is \[H=50+y_{max}=111.22\quad {\rm m}\] 

For further reading about uniform motion along the horizontal direction, see  speed, velocity, and acceleration problems .

Problem (3): A person standing on the edge of a 50-m-high cliff throws a stone horizontally at a speed of 18 m/s.  (a) What is the initial position of the stone?  (b) What are the components of the initial velocity? (c) What are the $x$- and $y$-components of the velocity of the stone at any arbitrary time $t$? (d) How long will it take the stone to strike the bottom of the cliff? (e) With what angle and speed does the stone strike the ground below the cliff?

Solution : As mentioned repeatedly, as a first step to solving a projectile motion problem, choose a relevant coordinate system. 

(a) Usually, we place the origin of the coordinate system at the point where the projectile is thrown. In this case, the coordinate of the initial position is $x_0=0\, , \, y_0=0$.

If we had chosen the coordinate at the base of the cliff and placed the origin at that point, the position of the initial point would have been $x_0=0\, , \, y_0=50\,\rm m$. 

(b) The stone is thrown horizontally with $\theta=0$, so there is no vertical velocity component. Consequently, its initial speed components are $v_{0x}=18\,\rm m/s$ and $v_{0y}=0$. 

(c) Remember that the velocity components of a projectile change over time according to the following formula. Substituting the given values into it, we get:  \begin{align*} v_x&=v_{0x}=18\,\rm m/s \\\\ v_y&=v_{0y}-gt\\&=-9.8t \end{align*}  (d) If we take the top of the cliff to be the origin of our coordinate system, with  $x_0=y_0=0$, the stone hits the ground $50\,\rm m$ below our chosen origin. Therefore, the coordinates of that point would be $x_0=?\, ,\, y=-50\,\rm m$. 

Use the vertical displacement kinematic equation below, substitute the numerical values into that, and solve for time $t$ get \begin{gather*} y-y_0=-\frac 12 gt^2+v_{0y}t \\\\ -50-0=-\frac 12 (9.8)t^2-0 \\\\ \Rightarrow \quad \boxed{t=3.2\,\rm s} \end{gather*}  (e) First, find the velocity components just before the stone hits the ground. We know that it takes about $3.2\,\rm s$ for the stone to reach the bottom of the cliff. Thus, put this time value into the formulas of velocity components at any time $t$. \begin{align*} v_x&=v_{0x}=18\,\rm m/s \\\\ v_y&=v_{0y}-gt\\&=-9.8\times 3.2 \\&=-31.36\,\rm m/s \end{align*} Notice that the negative sign here indicates that the vertical velocity component just before hitting the ground points downward, as expected. 

Recall from the section on  vector practice problems that when we have the components of a vector, we can find its magnitude using the Pythagorean theorem. This gives us: \begin{align*} v&=\sqrt{v_x^2+v_y^2} \\\\ &=\sqrt{18^2+(-31.36)^2} \\ &=\boxed{36.15\,\rm m/s} \end{align*} We can also find the angle of impact with the ground using the velocity components: \begin{align*} \tan\theta&=\frac{v_y}{v_x} \\\\ &=\frac{-31.36}{18} \\\\ &=-1.74 \end{align*} Taking the inverse tangent of both sides gives: \[\theta=\tan^{-1}(-1.74)=\boxed{-60.1^\circ} \] Therefore, the stone hit the ground at an angle of about $60^\circ$ with a speed of $36.15\,\rm m/s$. The negative angle indicates that it is below horizontal, which is to be expected.

Problem (4): A book slides off a frictionless tabletop with a constant speed of 1.1 m/s, and 0.35 seconds later strikes the floor.  (a) How high is the tabletop from the floor? (b) What is the horizontal distance to that point where the book strikes the floor? (c) How fast and at what angle does the book strike the floor? 

Solution : The total time the book is in the air is $t=0.35\,\rm s$, and the initial speed with which the book leaves the table horizontally is $v_{0x}=1.1\,\rm m/s$, and $v_{0y}=0$. 

Contrary to the earlier question, in this case, for practice, we place the origin at the bottom of the tabletop. Thus, the coordinates of the initial position would be $x_0=0\, ,\, y_0=h=?$ and the coordinates of the landing point would be $x=? \, , \, y=0$. 

(a) If we substitute the given values into the vertical displacement kinematic equation, we will have \begin{gather*} y-y_0=-\frac 12 gt^2+v_{0y}t \\\\ 0-h=-\frac 12 (9.8)(1.1)^2+0 \\\\ \Rightarrow \quad \boxed{h=5.93\,\rm m} \end{gather*} Thus, the table is 3-m-high.

(b)  Remember that projectile motion involves two motions that are uniformly accelerated in the horizontal and vertical directions.  The time taken by the projectile in the vertical direction is equal to the horizontal time to the landing point.  

Between the start and final points in the horizontal direction of the projectile path, we can write $\Delta x=v_{0x}t$. Substituting the given values into is get \begin{gather*} x-0=1.1\times (0.35) \\ \Rightarrow \quad \boxed{x=0.385\,\rm m} \end{gather*} where $\Delta x=x-x_0$. Therefore, the book strikes the floor about $39\,\rm cm$ ahead of the base of the table. 

(c) This part is straightforward. Substitute the given numerical values into the velocity vector component formulas as below to find the velocity components just before the book strikes the floor. \begin{align*} v_x&=v_{0x}=1.1\,\rm m/s \\\\ v_y&=v_{0y}-gt \\\\ &=0-(9.8)(0.35) \\&=-3.43\,\rm m/s \end{align*} The square root of the sum of the velocity components squared gives the velocity of the book at the instant of hitting \begin{align*} v&=\sqrt{v_x^2+v_y^2} \\\\ &=\sqrt{1.1^2+(-3.43)^2} \\\\ &=\boxed{3.6\,\rm m/s} \end{align*}  and angle of impact with the ground is found as below \begin{align*} \theta&=\tan^{-1}\left(\frac{v_y}{v_x}\right) \\\\ &=\tan^{-1}\left(\frac{-3.43}{1.1}\right) \\\\ &=\tan^{-1}(-3.11) \\\\&=\boxed{-72.17^\circ} \end{align*} Overall, the book hit the ground at an angle of about $72^\circ$ with a speed of nearly $3.6\,\rm m/s$. 

Problem (5): A cannonball is fired from a cliff with a speed of 800 m/s at an angle of 30° below the horizontal. How long will it take to reach 150 m below the firing point? 

Solution: First, choose the origin to be at the firing point, so $x_0=y_0=0$. Now, list the known values as follows      (i) Projectile's initial velocity = $800\,{\rm m/s}$      (ii) Angle of projectile: $-30^\circ$, the negative sign is because the throw is below the horizontal.     (iii) y-coordinate of the final point, 150 m below the origin, $y=-150\,{\rm m}$.       In this problem, the time it takes for the cannonball to reach 100 m below the starting point is required. 

Since the displacement to that point is known, we apply the vertical displacement projectile formula to find the needed time as below \begin{align*} y-y_0&=-\frac 12 gt^2+(v_0\sin\theta)t\\\\-150&=-\frac 12 (9.8)t^2+(800)\sin(-30^\circ)t\\\\ \Rightarrow & \ 4.9t^2+400t-150=0\end{align*} The above quadratic equation has two solutions, $t_1=0.37\,{\rm s}$ and $t_2=-82\,{\rm s}$. It is obvious that the second time is not acceptable.

Therefore, the cannonball takes 0.37 seconds to reach 150 meters below the firing point. 

Problem (6): Someone throws a stone into the air from ground level. The horizontal component of velocity is 25 m/s and it takes 3 seconds for the stone to come back to the same height as before. Find  (a) The range of the stone. (b) The initial vertical component of velocity (c) The angle of projection. 

Solution: The known values are      (i) The initial horizontal component of velocity, $v_{0x}=25\,{\rm m/s}$.      (ii) The time between the initial and final points, which are at the same level, $t=3\,{\rm s}$. 

(a)  The range of projectile motion is defined as the horizontal distance between the launch point and impact at the same elevation.

Because the horizontal motion in projectiles is a motion with constant velocity, the distance traveled in this direction is obtained as $x=v_{0x}t$, where $v_{0x}$ is the initial component of the velocity. 

If you put the total time the projectile is in the air into this formula, you get the range of the projectile. 

In this problem, the stone is thrown from the ground level and, after 3 seconds, reaches the same height. Thus, this is the total time of the projectile. 

Hence, the range of the stone is found as below \begin{align*} x&=v_{0x}t\\&=(25)(3)\\&=75\,{\rm m}\end{align*}

(b)  The initial vertical component of the projectile's velocity appears in two equations, $v_y=v_{0y}-gt$ and $y-y_0=-\frac 12 gt^2+v_{0y}t$. 

Using the second formula is more straightforward because the stone reaches the same height, so its vertical displacement between the initial and final points is zero, i.e., $y-y_0=0$. Setting this into the vertical distance projectile equation, we get \begin{align*} y-y_0&=-\frac 12 gt^2+v_{0y}t\\\\ 0&=-\frac 12 (9.8)(3)^2+v_{0y}(3) \\\\ \Rightarrow v_{0y}&=14.7\quad{\rm m/s}\end{align*} To use the first formula, we need some extra facts about projectile motion in the absence of air resistance, as below      (i) The vertical velocity is zero at the highest point of the path of the projectile, i.e., $v_y=0$.     (ii) If the projectile lands at the same elevation from which it was launched, then the time it takes to reach the highest point of the trajectory is half the total time between the initial and final points.      The second note, in the absence of air resistance, is only valid. 

In this problem, the total flight time is 3 s because air resistance is negligible, so 1.5 seconds are needed for the stone to reach the maximum height of its path. 

Therefore, using the second equation, we can find $v_{0y}$ as below \begin{align*} v_y&=v_{0y}-gt\\0&=v_{0y}-(9.8)(1.5) \\\Rightarrow v_{0y}&=14.7\quad {\rm m/s}\end{align*}  (c) The projection angle is the angle at which the projectile is thrown into the air and performs a two-dimensional motion. 

Once the components of the initial velocity are available, using trigonometry, we can find the angle of projection as below \begin{align*} \theta&=\tan^{-1}\left(\frac{v_{0y}}{v_{0x}}\right)\\\\&=\tan^{-1}\left(\frac{14.7}{25}\right)\\\\&=+30.45^\circ\end{align*} Therefore, the stone is thrown into the air at an angle of about $30^\circ$ above the horizontal.

Problem (7): A ball is thrown at an angle of 60° with an initial speed of 200 m/s. (Neglect the air resistance.) (a) How long is the ball in the air? (b) Find the maximum horizontal distance traveled by the ball. (c) What is the maximum height reached by the ball?

Solution: We choose the origin to be the ball's initial position so that $x_0=y_0=0$. The given data is      (i) The projection angle: $60^\circ$.     (ii) Initial speed : $v_0=200\,{\rm m/s}$. 

(a) The initial and final points of the ball are at the same level, i.e., $y-y_0=0$. 

Thus, the total time the ball is in the air is found by setting $y=0$ in the projectile equation $y=-\frac 12 gt^2+v_{0y}t$ and solving for time $t$ as below \begin{align*} y&=-\frac 12 gt^2+(v_0\sin\theta)t\\\\0&=-\frac 12 (9.8)t^2+(200)(\sin 60^\circ)t\\\\\Rightarrow \quad & \boxed{(-4.9t+100\sqrt{3})t=0} \end{align*} The above expression has two solutions for $t$. One is the initial time, $t_1=0$, and the other is computed as $t_2=35.4\,{\rm s}$. 

Hence, the ball has been in the air for about 35 s. 

(b) The horizontal distance is called the range of the projectile. By inserting the above time (total flight time) into the horizontal distance projectile equation $x=v_{0x}t$, we can find the desired distance traveled. \begin{align*} x&=(v_0\cos\theta)t\\&=(200)(\cos 60^\circ)(35.4)\\&=3540 \quad {\rm m}\end{align*} Therefore, the ball hits the ground 3540 meters away from the throwing point.

(c) Using the projectile equation $v_y^2-v_{0y}^2=-2g(y-y_0)$, setting $v_y=0$ at the highest point of the path, and solving for the vertical distance $y$, the maximum height is found as follows \begin{align*} v_y^2-v_{0y}^2&=-2g(y-y_0)\\0-(200\sin 60^\circ)^2&=-2(9.8)y\\\Rightarrow y&=1531\quad {\rm m}\end{align*} Another method: As mentioned above, the ball hits the ground at the same level as before, so by knowing the total flight time and halving it, we can find the time it takes the ball to reach the highest point of its trajectory. 

Therefore, by setting the half of the total flight time in the following projectile kinematic formula and solving for $y$, we can find the maximum height as \begin{align*} y-y_0&=-\frac 12 gt^2+(v_0\sin\theta)t\\\\y&=-\frac 12 (9.8)(17.7)^2+(200\sin60^\circ)(17.7)\\\\&=1531\quad {\rm m}\end{align*} Hence, the ball reaches 1531 meters above the launch point. 

Problem (8): What are the horizontal range and maximum height of a bullet fired with a speed of 20 m/s at 30° above the horizontal?

Solution : first, find the total flight time, then insert it into the horizontal displacement projectile equation $x=v_{0x}t$ to find the range of the bullet.

Because the bullet lands at the same level as the original, its vertical displacement is zero, $y-y_0=0$, in the following projectile formula we can find the total flight time \begin{align*} y-y_0&=-\frac 12 gt^2+(v_0\sin\theta)t\\\\0&=-\frac 12 (9.8)t^2+(20)(\sin30^\circ)t\\\\ \Rightarrow & (-4.9t+10)t=0\end{align*} Solving for time $t$ gives two solutions: one is the initial time $t_1=0$, and the other is $t_2=1.02\,{\rm s}$. Thus the total time of flight is 2.04 s. 

Therefore, the maximum horizontal distance traveled by the bullet, which is defined as the range of the projectile, is calculated as \begin{align*} x&=(v_0\cos\theta)t\\&=(20\cos 30^\circ)(2.04)\\&=35.3\quad {\rm m}\end{align*} Hence, the bullet lands about 17 m away from the launch point.

Because the air resistance is negligible and the bullet lands at the same height as the original, the time it takes to reach the highest point of its path is always half the total flight time. 

On the other hand, recall that the vertical component of velocity at the maximum height is always zero, i.e., $v_y=0$. By inserting these two notes into the following projectile equation, we have \begin{align*} y&=-\frac 12 gt^2+(v_0\sin\theta)t\\\\&=-\frac 12 (9.8)(1.02)^2+(20\sin 30^\circ)(1.02)\\\\&=5.1\quad {\rm m}\end{align*} We could also use the kinematic equation $v_y^2-v_{0y}^2=-2g(y-y_0)$, to find the maximum height as below \begin{align*} v_y^2-(v_0 \sin \theta)^2 &=-2g(y-y_0) \\ 0-(20\sin 30^\circ)^2 &=-2(9.8)H\\ \Rightarrow H&= 5.1\quad {\rm m} \end{align*} I think the second method is much simpler. 

Problem (9): A projectile is fired horizontally at a speed of 8 m/s from an 80-m-high cliff. Find  (a) The velocity just before the projectile hits the ground.  (b) The angle of impact.

Solution : In this problem, the angle of the projectile is zero because it is fired horizontally. 

The velocity at each point of a projectile trajectory (path) is obtained by the following formula: \[v=\sqrt{v_x^2+v_y^2}\] where $v_x$ and $v_y$ are the horizontal and vertical components of the projectile's velocity at any instant of time. 

(a) Recall that the horizontal component of the projectile's velocity is always constant and, for this problem, is found as \begin{align*} v_x&=v_0\cos\theta\\&=8\times \cos 0^\circ\\&=8\quad {\rm m/s}\end{align*} To find the vertical component of the projectile velocity at any moment, $v_y=v_0\sin\theta-gt$, we should find the time taken to reach that point. 

In this problem, that point is located just before striking the ground, whose coordinates are $y=-80\,{\rm m}, x=?$.

Because it is below the origin, which is assumed to be at the firing point, we inserted a minus sign. 

Because displacement in the vertical direction is known, we can use the projectile formula for vertical distance. 

By setting $y=-80$ into it and solving for the time $t$ needed the projectile reaches the ground, we get \begin{align*} y&=-\frac 12 gt^2+(v_0\sin\theta)t\\\\-80&=-\frac 12 (9.8)t^2+(8\times \sin 0^\circ)t\\\\\Rightarrow t&=\sqrt{\frac{2(80)}{9.8}}\\\\&=2.86\quad {\rm s}\end{align*} Now insert this time into the $y$-component of the projectile's velocity to find $v_y$ just before hitting the ground \begin{align*} v_y&=v_0\sin\theta-gt\\ &=8\sin 0^\circ-(9.8)(2.86)\\&=-28\quad {\rm m/s}\end{align*} Now that both components of the velocity are available, we can compute its magnitude as below \begin{align*} v&=\sqrt{v_x^2+v_y^2}\\\\&=\sqrt{8^2+(-28)^2}\\\\&=29.1\quad {\rm m/s}\end{align*} Therefore, the projectile hits the ground at a speed of 29.1 m/s. 

(b) At any instant of time, the velocity of the projectile makes some angle with the horizontal, whose magnitude is obtained as the follows: \[\alpha=\tan^{-1}\left(\frac{v_y}{v_x}\right)\] Substituting the above values into this formula, we get \[\alpha =\tan^{-1}\left(\frac{-28}{8}\right)=-74^\circ\] Therefore, the projectile hits the ground at an angle of 74° below the horizontal. 

Problem (10): From a cliff 100 m high, a ball is kicked at $30^\circ$ above the horizontal with a speed of $20\,{\rm m/s}$. How far from the base of the cliff did the ball hit the ground? (Assume $g=10\,{\rm m/s^2}$).  

Solution: Again, similar to any projectile motion problem, we first select a coordinate system and then draw the path of the projectile as shown in the figure below, 

We choose the origin to be at the kicking point above the cliff, setting $x_0=y_0=0$ in the kinematic equations. 

The coordinate of the hitting point to the ground is $y=-100\,{\rm m}, x = ?$. A negative is inserted because the final point is below the origin.

Now, we find the common quantity in projectile motions—that is, the time between the initial and final points, called total flight time.

To find the total time the ball was in the air, we can use the vertical equation and solve for the unknown $t$ as follows \begin{align*} y&=-\frac 12 gt^2 +(v_0\sin \theta)t \\\\ -100&=-\frac 12 (10) t^2+(20\sin 30^\circ)t\\\\&\Rightarrow \quad \boxed{t^2-2t-20=0} \end{align*} The solutions of a quadratic equation $at^2+bt+c=0$ are found by the formula below \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] By matching the above constant coefficients with the standard quadratic equation, we can find the total time as below \[t=\frac{-(-2)\pm\sqrt{(-2)^2-4(1)(-20)}}{2(1)}\] After simplifying, two solutions are obtained, $t_1=5.6\,{\rm s}$ and $t_2=-3.6\,{\rm s}$. 

It is obvious that time cannot be negative in physics, so the acceptable answer is $t_1$.

It is the time it takes the ball to travel in the vertical direction. On the other hand, it is also the time it takes the ball to travel the horizontal distance between kicking and hitting points. 

We insert this time into the horizontal equation to find the horizontal distance traveled, known as the projectile's range. \begin{align*} x&=(v_0\cos \theta)t\\ &=(20)(\cos 30^\circ)(5.6)\\&=97\quad {\rm m}\end{align*} 

Problem (11): A cannonball is fired from the top of a $200\,{\rm m}$ cliff with a speed of $60\,{\rm m/s}$ at an angle of $60^\circ$ above the horizontal. Where does the ball land closest to which of the following choices? (Neglect air resistance.) (a) 90m        (b) 402 m        (c) 151 m     (d) 200 m

Solution : The horizontal distance from the point where the cannonball fired is wanted, $x=?$. The appropriate kinematic equation for this is \[x=(v_0\cos\theta)t\] The only thing we need is the total time the ball is in the air before landing.

If we take the firing point as the origin, then the cannonball lands $200\,{\rm m}$ below the origin. So, the ball's vertical displacement is $\Delta y=-200\,{\rm m}$. Set this in the vertical displacement equation, $\Delta y=-\frac 12 gt^2+(v_0\sin\theta)t$, and solve for $t$. \begin{gather*} \Delta y=-\frac 12 gt^2 +(v_0\sin\theta)t \\\\ -200=-\frac 12 (10)t^2 +(60\sin 60^\circ)t \\\\  \Rightarrow\quad 5t^2-30\sqrt{3}t-200=0 \\\\ \boxed{t^2-6\sqrt{3}t-40=0}\end{gather*} where we used $\sin 60^\circ=\frac{\sqrt{3}}2$. Using a graphing calculator plot this equation and find its intersections with the horizontal. Consequently, the total flight time is \[\boxed{t=13.4\,{\rm m}}\] Now, substitute this into the horizontal displacement projectile equation below \begin{align*} x&=(v_0\cos\theta)t \\\\ &=(60\,\cos 60^\circ)(13.4) \\\\ &=402\,{\rm m}\end{align*} Thus, the correct answer is (b).

Problem (12): During a kicking drill, a soccer player kicks two successive balls at different angles but with exactly the same speed. The paths that the balls follow are depicted in the figure shown. Which ball remains in the air for a longer time? Neglect air resistance.

Solution : both balls land at the same level (or the ground), so the total vertical displacement of both is the same, $\Delta y_1=\Delta y_2=0$. The initial kicking speeds are equal $v_{01}=v_{02}$ and as seen from the figure, ball $1$ has greater kicking angle than ball $2$, i.e., $\theta_1>\theta_2$. Setting $\Delta y=0$ in the following equation and solving for $t$, gives us the total time the ball is in the air \begin{gather*} \Delta y=-\frac 12 gt^2+(v_0 \sin\theta)t \\\\ 0=-\frac 12 gt^2+(v_0 \sin\theta)t \\\\ \Rightarrow \boxed{t=\frac{2v_0 \sin\theta}{g}} \end{gather*} Now, substitute the given data into the equation to compare the flight time of both balls \begin{gather*} \theta_1>\theta_2 \\ \sin\theta_1 > \sin\theta_2 \\\\ \frac{2v_0 \sin\theta_1}{g} > \frac{2v_0 \sin\theta_2}{g} \\\\ \Rightarrow \boxed{t_1>t_2} \end{gather*} Thus, ball $1$ will be in the air for a longer time. 

Author : Dr. Ali Nemati Date Published: 6/15/2021

Updated:  Jun 22, 2023

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Projectile Motion Example Problem – Physics Homework Help 1

Throwing or shooting a projectile follows a parabolic course. If you know the initial velocity and angle of elevation of the projectile, you can find its time aloft, maximum height or range. You can also its altitude and distance travelled if given a time. This example problem shows how to do all of these.

Projectile Motion Example Problem: A cannon is fired with muzzle velocity of 150 m/s at an angle of elevation = 45°. Gravity = 9.8 m/s 2 . a) What is the maximum height the projectile reaches? b) What is the total time aloft? c) How far away did the projectile land? (Range) d) Where is the projectile at 10 seconds after firing?

Let’s set up what we know. First, let’s define our variables.

V 0 = initial velocity = muzzle velocity = 150 m/s v x = horizontal velocity component v y = vertical velocity component θ = angle of elevation = 45° h = maximum height R = range x = horizontal position at t=10 s y = vertical position at t=10 s m = mass of projectile g = acceleration due to gravity = 9.8 m/s 2

Part a) Find h.

The formulas we will be using are:

d = v 0 t + ½at 2

v f – v 0 = at

In order to find the distance h, we need to know two things: the velocity at h and the amount of time it takes to get there. The first is easy. The vertical component of the velocity is equal to zero at point h. This is the point where the upward motion is stopped and the projectile begins to fall back to Earth.

The initial vertical velocity is v 0y = v 0 ·sinθ v 0y = 150 m/s · sin(45°) v 0y = 106.1 m/s

Now we know the beginning and final velocity. The next thing we need is the acceleration.

The only force acting on the projectile is the force of gravity. Gravity has a magnitude of g and a direction in the negative y direction.

F = ma = -mg

solve for a

Now we have enough information to find the time. We know the initial vertical velocity (V 0y ) and the final vertical velocity at h (v hy = 0)

v hy – v 0y  = at 0 – v 0y = -9.8 m/s 2 ·t 0 – 106.1 m/s = -9.8 m/s 2 ·t

Solve for t

t = 10.8 s

Now solve the first equation for h

h = v 0y t + ½at 2 h = (106.1 m/s)(10.8 s) + ½(-9.8 m/s 2 )(10.8 s) 2 h = 1145.9 m – 571.5 m h = 574.4 m

The highest height the projectile reaches is 574.4 meters.

Part b: Find total time aloft.

We’ve already done most of the work to get this part of the question if you stop to think. The projectile’s trip can be broken into two parts: going up and coming down.

t total = t up + t down

The same acceleration force acts on the projectile in both directions. The time down takes the same amount of time it took to go up.

t up = t down

t total = 2 t up

we found t up in Part a of the problem: 10.8 seconds

t total = 2 (10.8 s) t total = 21.6 s

The total time aloft for the projectile is 21.6 seconds.

Part c: Find range R

To find the range, we need to know the initial velocity in the x direction.

v 0x = v 0 cosθ v 0x = 150 m/s·cos(45) v 0x = 106.1 m/s

To find the range R, use the equation:

R = v 0x t + ½at 2

There is no force acting along the x-axis. This means the acceleration in the x-direction is zero. The equation of motion is reduced to:

R = v 0x t + ½(0)t 2 R = v 0x t

The range is the point where the projectile strikes the ground which happens at the time we found in Part b of the problem.

R = 106.1 m/s · 21.6s R = 2291.8 m

The projectile landed 2291.8 meters from the canon.

Part d: Find the position at t = 10 seconds.

The position has two components: horizontal and vertical position. The horizontal position, x, is far downrange the projectile is after firing and the vertical component is the current altitude, y, of the projectile.

To find these positions, we will use the same equation:

First, let’s do the horizontal position. There is no acceleration in the horizontal direction so the second half of the equation is zero, just like in Part c.

We are given t = 10 seconds. V 0x was calculated in Part c of the problem.

x = 106.1 m/s · 10 s x = 1061 m

Now do the same thing for the vertical position.

y = v 0y t + ½at 2

We saw in Part b that v 0y = 109.6 m/s and a = -g = -9.8 m/s 2 . At t = 10 s:

y = 106.1 m/s · 10 s + ½(-9.8 m/s 2 )(10 s) 2 y = 1061 – 490 m y = 571 m

At t=10 seconds, the projectile is at (1061 m, 571 m) or 1061 m downrange and at an altitude of 571 meters.

If you need to know the velocity of the projectile at a specific time, you can use the formula

v – v 0  = at

and solve for v. Just remember velocity is a vector and will have both x and y components.

This specific example can be easily adapted for any initial velocity and any angle of elevation. If the cannon is fired on another planet with a different force of gravity, just change the value of g accordingly.

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Physics library

Course: physics library   >   unit 2.

• Horizontally launched projectile

What is 2D projectile motion?

• Visualizing vectors in 2 dimensions
• Projectile at an angle
• Launching and landing on different elevations
• Total displacement for projectile
• 2D projectile motion: Identifying graphs for projectiles
• 2D projectile motion: Vectors and comparing multiple trajectories
• What are velocity components?
• Unit vectors and engineering notation
• Unit vector notation
• Unit vector notation (part 2)

What's a 2D projectile?

How do we handle 2d projectile motion mathematically, horizontal direction:, vertical direction:, what's confusing about 2d projectile motion, what do solved examples involving 2d projectile motion look like, example 1: horizontally launched water balloon, example 2: pumpkin launched at an angle, want to join the conversation.

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