homework 3 free fall and graphs

3.5 Free Fall

Learning objectives.

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

• Use the kinematic equations with the variables y and g to analyze free-fall motion.
• Describe how the values of the position, velocity, and acceleration change during a free fall.
• Solve for the position, velocity, and acceleration as functions of time when an object is in a free fall.

An interesting application of Equation 3.4 through Equation 3.14 is called free fall , which describes the motion of an object falling in a gravitational field, such as near the surface of Earth or other celestial objects of planetary size. Let’s assume the body is falling in a straight line perpendicular to the surface, so its motion is one-dimensional. For example, we can estimate the depth of a vertical mine shaft by dropping a rock into it and listening for the rock to hit the bottom. But “falling,” in the context of free fall, does not necessarily imply the body is moving from a greater height to a lesser height. If a ball is thrown upward, the equations of free fall apply equally to its ascent as well as its descent.

The most remarkable and unexpected fact about falling objects is that if air resistance and friction are negligible, then in a given location all objects fall toward the center of Earth with the same constant acceleration , independent of their mass . This experimentally determined fact is unexpected because we are so accustomed to the effects of air resistance and friction that we expect light objects to fall slower than heavy ones. Until Galileo Galilei (1564–1642) proved otherwise, people believed that a heavier object has a greater acceleration in a free fall. We now know this is not the case. In the absence of air resistance, heavy objects arrive at the ground at the same time as lighter objects when dropped from the same height Figure 3.26 .

In the real world, air resistance can cause a lighter object to fall slower than a heavier object of the same size. A tennis ball reaches the ground after a baseball dropped at the same time. (It might be difficult to observe the difference if the height is not large.) Air resistance opposes the motion of an object through the air, and friction between objects—such as between clothes and a laundry chute or between a stone and a pool into which it is dropped—also opposes motion between them.

For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free fall . The force of gravity causes objects to fall toward the center of Earth. The acceleration of free-falling objects is therefore called acceleration due to gravity . Acceleration due to gravity is constant, which means we can apply the kinematic equations to any falling object where air resistance and friction are negligible. This opens to us a broad class of interesting situations.

Acceleration due to gravity is so important that its magnitude is given its own symbol, g . It is constant at any given location on Earth and has the average value

Although g varies from 9.78 m/s 2 to 9.83 m/s 2 , depending on latitude, altitude, underlying geological formations, and local topography, let’s use an average value of 9.8 m/s 2 rounded to two significant figures in this text unless specified otherwise. Neglecting these effects on the value of g as a result of position on Earth’s surface, as well as effects resulting from Earth’s rotation, we take the direction of acceleration due to gravity to be downward (toward the center of Earth). In fact, its direction defines what we call vertical. Note that whether acceleration a in the kinematic equations has the value + g or − g depends on how we define our coordinate system. If we define the upward direction as positive, then a = − g = −9.8 m/s 2 , a = − g = −9.8 m/s 2 , and if we define the downward direction as positive, then a = g = 9.8 m/s 2 a = g = 9.8 m/s 2 .

One-Dimensional Motion Involving Gravity

The best way to see the basic features of motion involving gravity is to start with the simplest situations and then progress toward more complex ones. So, we start by considering straight up-and-down motion with no air resistance or friction. These assumptions mean the velocity (if there is any) is vertical. If an object is dropped, we know the initial velocity is zero when in free fall. When the object has left contact with whatever held or threw it, the object is in free fall. When the object is thrown, it has the same initial speed in free fall as it did before it was released. When the object comes in contact with the ground or any other object, it is no longer in free fall and its acceleration of g is no longer valid. Under these circumstances, the motion is one-dimensional and has constant acceleration of magnitude g . We represent vertical displacement with the symbol y .

Kinematic Equations for Objects in Free Fall

We assume here that acceleration equals − g (with the positive direction upward).

Problem-Solving Strategy

• Decide on the sign of the acceleration of gravity. In Equation 3.15 through Equation 3.17 , acceleration g is negative, which says the positive direction is upward and the negative direction is downward. In some problems, it may be useful to have acceleration g as positive, indicating the positive direction is downward.
• Draw a sketch of the problem. This helps visualize the physics involved.
• Record the knowns and unknowns from the problem description. This helps devise a strategy for selecting the appropriate equations to solve the problem.
• Decide which of Equation 3.15 through Equation 3.17 are to be used to solve for the unknowns.

Example 3.14

Free fall of a ball.

• Substitute the given values into the equation: y = y 0 + v 0 t − 1 2 g t 2 − 98.0 m = 0 − ( 4.9 m/s ) t − 1 2 ( 9.8 m/s 2 ) t 2 . y = y 0 + v 0 t − 1 2 g t 2 − 98.0 m = 0 − ( 4.9 m/s ) t − 1 2 ( 9.8 m/s 2 ) t 2 . This simplifies to t 2 + t − 20 = 0 . t 2 + t − 20 = 0 . This is a quadratic equation with roots t = −5.0 s and t = 4.0 s t = −5.0 s and t = 4.0 s . The positive root is the one we are interested in, since time t = 0 t = 0 is the time when the ball is released at the top of the building. (The time t = −5.0 s t = −5.0 s represents the fact that a ball thrown upward from the ground would have been in the air for 5.0 s when it passed by the top of the building moving downward at 4.9 m/s.)
• Using Equation 3.15 , we have v = v 0 − g t = −4.9 m/s − ( 9.8 m/s 2 ) ( 4.0 s ) = −44.1 m/s . v = v 0 − g t = −4.9 m/s − ( 9.8 m/s 2 ) ( 4.0 s ) = −44.1 m/s .

Significance

Example 3.15, vertical motion of a baseball.

• Equation 3.16 gives y = y 0 + v 0 t − 1 2 g t 2 y = y 0 + v 0 t − 1 2 g t 2 0 = 0 + v 0 ( 5.0 s ) − 1 2 ( 9.8 m / s 2 ) ( 5.0 s ) 2 , 0 = 0 + v 0 ( 5.0 s ) − 1 2 ( 9.8 m / s 2 ) ( 5.0 s ) 2 , which gives v 0 = 24.5 m/s v 0 = 24.5 m/s .
• At the maximum height, v = 0 v = 0 . With v 0 = 24.5 m/s v 0 = 24.5 m/s , Equation 3.17 gives v 2 = v 0 2 − 2 g ( y − y 0 ) v 2 = v 0 2 − 2 g ( y − y 0 ) 0 = ( 24.5 m/s ) 2 − 2 ( 9.8 m/s 2 ) ( y − 0 ) 0 = ( 24.5 m/s ) 2 − 2 ( 9.8 m/s 2 ) ( y − 0 ) or y = 30.6 m . y = 30.6 m .
• To find the time when v = 0 v = 0 , we use Equation 3.15 : v = v 0 − g t v = v 0 − g t 0 = 24.5 m/s − ( 9.8 m/s 2 ) t . 0 = 24.5 m/s − ( 9.8 m/s 2 ) t . This gives t = 2.5 s t = 2.5 s . Since the ball rises for 2.5 s, the time to fall is 2.5 s.
• The acceleration is 9.8 m/s 2 everywhere, even when the velocity is zero at the top of the path. Although the velocity is zero at the top, it is changing at the rate of 9.8 m/s 2 downward.
• The velocity at t = 5.0 s t = 5.0 s can be determined with Equation 3.15 : v = v 0 − g t = 24.5 m/s − 9.8 m/s 2 ( 5.0 s ) = −24.5 m/s . v = v 0 − g t = 24.5 m/s − 9.8 m/s 2 ( 5.0 s ) = −24.5 m/s .

Check Your Understanding 3.7

A chunk of ice breaks off a glacier and falls 30.0 m before it hits the water. Assuming it falls freely (there is no air resistance), how long does it take to hit the water? Which quantity increases faster, the speed of the ice chunk or its distance traveled?

Example 3.16

Rocket booster.

• From Equation 3.17 , v 2 = v 0 2 − 2 g ( y − y 0 ) v 2 = v 0 2 − 2 g ( y − y 0 ) . With v = 0 and y 0 = 0 v = 0 and y 0 = 0 , we can solve for y : y = v 0 2 2 g = ( 2.0 × 10 2 m / s ) 2 2 ( 9.8 m / s 2 ) = 2040.8 m . y = v 0 2 2 g = ( 2.0 × 10 2 m / s ) 2 2 ( 9.8 m / s 2 ) = 2040.8 m . This solution gives the maximum height of the booster in our coordinate system, which has its origin at the point of release, so the maximum height of the booster is roughly 7.0 km.
• An altitude of 6.0 km corresponds to y = 1.0 × 10 3 m y = 1.0 × 10 3 m in the coordinate system we are using. The other initial conditions are y 0 = 0 , and v 0 = 200.0 m/s y 0 = 0 , and v 0 = 200.0 m/s . We have, from Equation 3.17 , v 2 = ( 200.0 m / s ) 2 − 2 ( 9.8 m / s 2 ) ( 1.0 × 10 3 m ) ⇒ v = ± 142.8 m / s . v 2 = ( 200.0 m / s ) 2 − 2 ( 9.8 m / s 2 ) ( 1.0 × 10 3 m ) ⇒ v = ± 142.8 m / s .

Interactive

Engage the Phet simulation below to learn about graphing polynomials. The shape of the curve changes as the constants are adjusted. View the curves for the individual terms (for example, y = bx ) to see how they add to generate the polynomial curve.

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Look at the given example below and try to understand what I tried to explain above.

Kinematics Exams and Solutions

acceleration due to gravity

Want to see an object accelerate?

• Pick something up with your hand and drop it. When you release it from your hand, its speed is zero. On the way down its speed increases. The longer it falls the faster it travels. Sounds like acceleration to me.
• But acceleration is more than just increasing speed. Pick up this same object and toss it vertically into the air. On the way up its speed will decrease until it stops and reverses direction. Decreasing speed is also considered acceleration.
• But acceleration is more than just changing speed. Pick up your battered object and launch it one last time. This time throw it horizontally and notice how its horizontal velocity gradually becomes more and more vertical. Since acceleration is the rate of change of velocity with time and velocity is a vector quantity, this change in direction is also considered acceleration.

In each of these examples the acceleration was the result of gravity. Your object was accelerating because gravity was pulling it down. Even the object tossed straight up is falling — and it begins falling the minute it leaves your hand. If it wasn't, it would have continued moving away from you in a straight line. This is the acceleration due to gravity .

What are the factors that affect this acceleration due to gravity? If you were to ask this of a typical person, they would most likely say "weight" by which they actually mean "mass" (more on this later). That is, heavy objects fall fast and light objects fall slow. Although this may seem true on first inspection, it doesn't answer my original question. "What are the factors that affect the acceleration due to gravity ?" Mass does not affect the acceleration due to gravity in any measurable way. The two quantities are independent of one another. Light objects accelerate more slowly than heavy objects only when forces other than gravity are also at work. When this happens, an object may be falling, but it is not in free fall. Free fall occurs whenever an object is acted upon by gravity alone.

Try this experiment.

• Obtain a piece of paper and a pencil. Hold them at the same height above a level surface and drop them simultaneously. The acceleration of the pencil is noticeably greater than the acceleration of the piece of paper, which flutters and drifts about on its way down.

Something else is getting in the way here — and that thing is air resistance (also known as aerodynamic drag). If we could somehow reduce this drag we'd have a real experiment. No problem.

• Repeat the experiment, but before you begin, wad the piece of paper up into the tightest ball possible. Now when the paper and pencil are released, it should be obvious that their accelerations are identical (or at least more similar than before).

We're getting closer to the essence of this problem. If only somehow we could eliminate air resistance altogether. The only way to do that is to drop the objects in a vacuum. It is possible to do this in the classroom with a vacuum pump and a sealed column of air. Under such conditions, a coin and a feather can be shown to accelerate at the same rate. (In the olden days in Great Britain, a coin called a guinea was used and so this demonstration is sometimes called the "guinea and feather".) A more dramatic demonstration was done on the surface of the moon — which is as close to a true vacuum as humans are likely to experience any time soon. Astronaut David Scott released a rock hammer and a falcon feather at the same time during the Apollo 15 lunar mission in 1971. In accordance with the theory I am about to present, the two objects landed on the lunar surface simultaneously (or nearly so). Only an object in free fall will experience a pure acceleration due to gravity.

the leaning tower of Pisa

Let's jump back in time for a bit. In the Western world prior to the 16th century, it was generally assumed that the acceleration of a falling body would be proportional to its mass — that is, a 10 kg object was expected to accelerate ten times faster than a 1 kg object. The ancient Greek philosopher Aristotle of Stagira (384–322 BCE), included this rule in what was perhaps the first book on mechanics. It was an immensely popular work among academicians and over the centuries it had acquired a certain devotion verging on the religious. It wasn't until the Italian scientist Galileo Galilei (1564–1642) came along that anyone put Aristotle's theories to the test. Unlike everyone else up to that point, Galileo actually tried to verify his own theories through experimentation and careful observation. He then combined the results of these experiments with mathematical analysis in a method that was totally new at the time, but is now generally recognized as the way science gets done. For the invention of this method, Galileo is generally regarded as the world's first scientist.

In a tale that may be apocryphal, Galileo (or an assistant, more likely) dropped two objects of unequal mass from the Leaning Tower of Pisa. Quite contrary to the teachings of Aristotle, the two objects struck the ground simultaneously (or very nearly so). Given the speed at which such a fall would occur, it is doubtful that Galileo could have extracted much information from this experiment. Most of his observations of falling bodies were really of round objects rolling down ramps. This slowed things down enough to the point where he was able to measure the time intervals with water clocks and his own pulse (stopwatches and photogates having not yet been invented). This he repeated "a full hundred times" until he had achieved "an accuracy such that the deviation between two observations never exceeded one-tenth of a pulse beat."

With results like that, you'd think the universities of Europe would have conferred upon Galileo their highest honor, but such was not the case. Professors at the time were appalled by Galileo's comparatively vulgar methods even going so far as to refuse to acknowledge that which anyone could see with their own eyes. In a move that any thinking person would now find ridiculous, Galileo's method of controlled observation was considered inferior to pure reason. Imagine that! I could say the sky was green and as long as I presented a better argument than anyone else, it would be accepted as fact contrary to the observation of nearly every sighted person on the planet.

Galileo called his method "new" and wrote a book called Discourses on Two New Sciences wherein he used the combination of experimental observation and mathematical reasoning to explain such things as one dimensional motion with constant acceleration, the acceleration due to gravity, the behavior of projectiles, the speed of light, the nature of infinity, the physics of music, and the strength of materials. His conclusions on the acceleration due to gravity were that…

the variation of speed in air between balls of gold, lead, copper, porphyry, and other heavy materials is so slight that in a fall of 100 cubits a ball of gold would surely not outstrip one of copper by as much as four fingers. Having observed this I came to the conclusion that in a medium totally devoid of resistance all bodies would fall with the same speed. For I think no one believes that swimming or flying can be accomplished in a manner simpler or easier than that instinctively employed by fishes and birds. When, therefore, I observe a stone initially at rest falling from an elevated position and continually acquiring new increments of speed, why should I not believe that such increases take place in a manner which is exceedingly simple and rather obvious to everybody? I greatly doubt that Aristotle ever tested by experiment. Galileo Galilei, 1638

Despite that last quote, Galileo was not immune to using reason as a means to validate his hypothesis. In essence, his argument ran as follows. Imagine two rocks, one large and one small. Since they are of unequal mass they will accelerate at different rates — the large rock will accelerate faster than the small rock. Now place the small rock on top of the large rock. What will happen? According to Aristotle, the large rock will rush away from the small rock. What if we reverse the order and place the small rock below the large rock? It seems we should reason that two objects together should have a lower acceleration . The small rock would get in the way and slow the large rock down. But two objects together are heavier than either by itself and so we should also reason that they will have a greater acceleration . This is a contradiction.

Here's another thought problem. Take two objects of equal mass. According to Aristotle, they should accelerate at the same rate. Now tie them together with a light piece of string. Together, they should have twice their original acceleration. But how do they know to do this? How do inanimate objects know that they are connected? Let's extend the problem. Isn't every heavy object merely an assembly of lighter parts stuck together? How can a collection of light parts, each moving with a small acceleration, suddenly accelerate rapidly once joined? We've argued Aristotle into a corner. The acceleration due to gravity is independent of mass.

Galileo made plenty of measurements related to the acceleration due to gravity but never once calculated its value (or if he did, I have never seen it reported anywhere). Instead he stated his findings as a set of proportions and geometric relationships — lots of them. His description of constant speed required one definition, four axioms, and six theorems. All of these relationships can now be written as the single equation in modern notation.

Algebraic symbols can contain as much information as several sentences of text, which is why they are used. Contrary to the common wisdom, mathematics makes life easier.

location, location, location

The generally accepted value for the acceleration due to gravity on and near the surface of the Earth is…

g  =  9.8 m/s 2

or in non-SI units…

g  =  35 kph/s = 22 mph/s = 32 feet/s 2

It is useful to memorize this number (as millions of people around the globe already have), however, it should also be pointed out that this number is not a constant . Although mass has no effect on the acceleration due to gravity, there are three factors that do. They are location, location, location.

Everyone reading this should be familiar with the images of the astronauts hopping about on the moon and should know that the gravity there is weaker than it is on the Earth — about one sixth as strong or 1.6 m/s 2 . That's why the astronauts were able to hop around on the surface easily despite the weight of their space suits. In contrast, gravity on Jupiter is stronger than it is on Earth — about two and a half times stronger or 25 m/s 2 . Astronauts cruising through the top of Jupiter's thick atmosphere would find themselves struggling to stand up inside their space ship.

On the Earth, gravity varies with latitude and altitude (to be discussed in a later chapter ). The acceleration due to gravity is greater at the poles than at the equator and greater at sea level than atop Mount Everest. There are also local variations that depend upon geology. The value of 9.8 m/s 2 — with only two significant digits — is true for all places on the surface of the Earth and holds for altitudes up to +10 km (the altitude of commercial jet airplanes) and depths down to −20 km (far below the deepest mines).

How crazy are you for accuracy? For most applications, the value of 9.8 m/s 2 is more than sufficient. If you're in a hurry, or don't have access to a calculator, or just don't need to be that accurate; rounding g on Earth to 10 m/s 2 is often acceptable. During a multiple choice exam where calculators aren't allowed, this is often the way to go. If you need greater accuracy, consult a comprehensive reference work to find the accepted value for your latitude and altitude.

If that's not good enough, then obtain the required instruments and measure the local value to as many significant digits as you can. You may learn something interesting about your location. I once met a geologist whose job it was to measure g across a portion of West Africa. When I asked him who he worked for and why he was doing this, he basically refused to answer other than to say that one could infer the interior structure of the Earth from a gravimetric map prepared from his findings. Knowing this, one might then be able to identify structures where valuable minerals or petroleum might be found.

Like all professions, those in the gravity measuring business ( gravimetry ) have their own special jargon. The SI unit of acceleration is the meter per second squared [m/s 2 ]. Split that into a hundred parts and you get the centimeter per second squared [cm/s 2 ] also known as the gal [Gal] in honor of Galileo. Note that the word for the unit is all lowercase, but the symbol is capitalized. The gal is an example of a Gaussian unit .

00 1 Gal = 1 cm/s 2  = 0.01 m/s 2 100 Gal = 100 cm/s 2  = 1 m/s 2 .

Split a gal into a thousand parts and you get a milligal [mGal].

1 mGal = 0.001 Gal  = 10 −5  m/s 2

Since Earth's gravity produces a surface acceleration of about 10 m/s 2 , a milligal is about 1 millionth of the value we're all used to.

1 g ≈ 10 m/s 2  = 1,000 Gal = 1,000,000 mGal

Measurements with this precision can be used to study changes in the Earth's crust, sea levels, ocean currents, polar ice, and groundwater. Push it a little bit further and it's even possible to measure changes in the distribution of mass in the atmosphere. Gravity is a weighty subject that will be discussed in more detail later in this book.

Don't confuse the phenomenon of acceleration due to gravity with the unit of a similar name. The quantity g has a value that depends on location and is approximately …

almost everywhere on the surface of the Earth. The unit g has the exact value of…

g =  9.80665 m/s 2

by definition.

They also use slightly different symbols. The defined unit uses the roman or upright g while the natural phenomenon that varies with location uses the italic or oblique g . Don't confuse g with g .

As mentioned earlier, the value of 9.8 m/s 2 with only two significant digits is valid for most of the surface of the Earth up to the altitude of commercial jet airliners, which is why it is used throughout this book. The value of 9.80665 m/s 2 with six significant digits is the so called standard acceleration due to gravity or standard gravity . It's a value that works for latitudes around 45° and altitudes not too far above sea level. It's approximately the value for the acceleration due to gravity in Paris, France — the hometown of the International Bureau of Weights and Measures . The original idea was to establish a standard value for gravity so that units of mass, weight, and pressure could be related — a set of definitions that are now obsolete. The Bureau chose to make this definition work for where their laboratory was located. The old unit definitions died out, but the value of standard gravity lives on. Now it's just an agreed upon value for making comparisons. It's a value close to what we experience in our everyday lives — just with way too much precision.

Some books recommend a compromise precision of 9.81 m/s 2 with three significant digits for calculations, but this book does not. At my location in New York City, the acceleration due to gravity is 9.80 m/s 2 . Rounding standard gravity to 9.81 m/s 2 is wrong for my location. The same is true all the way south to the equator where gravity is 9.780 m/s 2 at sea level — 9.81 m/s 2 is just too big. Head north of NYC and gravity gets closer and closer to 9.81 m/s 2 until eventually it is. This is great for Canadians in southern Quebec, but gravity keeps keeps increasing as you head further north. At the North Pole (and the South Pole too) gravity is a whopping 9.832 m/s 2 . The value 9.806 m/s 2 is midway between these two extremes, so it's sort of true to say that…

g  =  9.806 ± 0.026 m/s 2

This is not the same thing as an average, however. For that, use this value that someone else derived…

g  =  9.798 m/s 2

Here are my suggestions. Use the value of 9.8 m/s 2 with two significant digits for calculations on the surface of the Earth unless a value of gravity is otherwise specified. That seems reasonable. Use the value of 9.80665 m/s 2 with six significant digits only when you want to convert m/s 2 to g. That is the law.

The unit g is often used to measure the acceleration of a reference frame . This is technical language that will be elaborated upon later in another section of this book, but I will explain it with examples for now. As I write this, I'm sitting in front of my computer in my home office. Gravity is drawing my body down into my office chair, my arms toward the desk, and my fingers toward the keyboard. This is the normal 1 g (one gee) world we're all accustomed to. I could take a laptop computer with me to an amusement park, get on a roller coaster, and try to get some writing done there. Gravity works on a roller coaster just as it does at home, but since the roller coaster is accelerating up and down (not to mention side to side) the sensation of normal Earth gravity is lost. There will be times when I feel heavier than normal and times when I fell lighter than normal. These correspond to periods of more than one g and less than one g. I could also take my laptop with me on a trip to outer space. After a brief period of 2 or 3 g (two or three gees) accelerating away from the surface of the Earth, most space journeys are spent in conditions of apparent weightlessness or 0 g (zero gee). This happens not because gravity stops working (gravity has infinite range and is never repulsive), but because a spacecraft is an accelerating reference frame. As I said earlier, this concept will be discussed more thoroughly in a later section of this book.

Free Fall Motion: Explanation, Review, and Examples

• The Albert Team
• Last Updated On: February 16, 2023

Free fall and projectile motion describe objects that are moving through the air and acted on only by gravity. In this post, we will describe this type of motion using both graphs and kinematic equations. Since projectile motion involves two dimensions, these problems can be complex. We will explain many examples so you can see how to solve different types of projectile motion. 

What We Review

An object that is moving under only the influence of gravity is in free fall. In order for an object to be in free fall, wind and air resistance must be ignored. On Earth, all objects in free fall accelerate downward at the rate of gravity or 9.81\text{ m/s}^2 .

Applying Free Fall to Kinematic Equations

When analyzing free fall motion, we can apply the same kinematic equations as we did for motion on the ground. We can then use these equations to determine properties such as distance, time, and velocity. 

How to Find Distance Fallen for an Object in Free Fall

If an object is in free fall, we can use kinematic equations to find the distance it falls during a certain time. You will typically use the following kinematic equation to calculate the distance fallen:

In order to use this equation, you need to know the initial velocity of the object and the time of flight. Remember that the acceleration of a free falling object is always equal to the acceleration due to gravity, 9.81\text{ m/s}^2 . 

Many free fall physics problems will include scenarios where objects are dropped from rest. In this case, the initial velocity is zero and the first term of the kinematic equation above will cancel out. 

If the time is not known, another method for calculating the distance fallen is to use the following kinematic equation:

In this case, you must know the final velocity v_f of the object. Then, you can solve the equation for the distance d .

How to Find Time for an Object in Free Fall

The amount of time an object is in free fall will depend on its velocity and the distance it falls. Similar to distance, there are two equations you can use to find the time, depending on what you know. 

If you know the initial and final velocity of the object, then the simplest way to calculate time is using the kinematic equation:

This equation can be solved for time. Then, you’ll only need to substitute the values for the velocities and the acceleration due to gravity.

Another method to find time if you do not know the object’s final velocity is to use the equation:

Note that in this equation there are two terms that include the time t . Unless the initial velocity is zero, this can make it more challenging to solve this equation for time. If using this equation, you may need to use the quadratic formula to solve for time.

How to Find Final Velocity for an Object in Free Fall

The final velocity of an object in free fall depends on the amount of time it falls. Due to the acceleration of gravity, the velocity will increase every second by 9.81\text{ m/s} . The final velocity can be calculated using the equation:

If you do not know the amount of time the object is falling, another method for calculating the final velocity is using the kinematic equation: 

This equation requires that you instead know the distance that the object falls. If you are using this equation to find the final velocity, remember that the final velocity is squared in this equation. That means you will need to take a square root as your final step to solve for the final velocity. 

Examples of Free Fall

In this next section, we’ll apply the methods you just learned to solve some problems about free fall motion.

Example 1: How to Find the Distance for an Object Dropped from Rest

For example, an object is dropped from rest from the top of a tall building. It hits the ground 5\text{ s} after it is dropped. What is the height of the building? 

In this scenario, we know that the object’s initial velocity is zero because it was dropped from rest. We also know that the acceleration is 9.81\text{ m/s}^2 . This problem is asking us to find the distance the object falls. This will be equal to the height of the building.

Based on this information, we can use the following kinematic equation to find the distance:

Substituting the given values produces:

Therefore, the height of the building is about 123\text{ m} .

Example 2: How to Find the Final Velocity for an Object with Initial Velocity

In another example, an object in free fall has an initial, downward velocity of 2\text{ m/s} and falls a distance of 45\text{ m} . What is the object’s final velocity? 

In this scenario, we are given the object’s initial velocity, v_i and the distance d . We also know that the acceleration is 9.81\text{ m/s}^2 . Based on this information, we can use the following kinematic equation to find the final velocity:

Since the initial velocity is in the same direction as the acceleration (downward) we can use the same sign for both values.

Our last step is to eliminate the square by taking the square root:

Therefore, the final velocity of the object is about 30\text{ m/s} .

Motion Graphs for Objects in Free Fall

In addition to using physics equations, we can also represent free fall motion with motion graphs. Position-time graphs, velocity-time graphs, and acceleration-time graphs can tell us a lot about the object’s motion over time. Want a more in-depth review of motion graphs? Check out this blog post !

Position-Time Graph for an Object in Free Fall

In terms of position, many objects in free fall start at a high position, or height off the ground, and move downward. Objects in free fall accelerate due to gravity. Therefore, the position-time graph for free fall motion must be curved. This means that objects in free fall start with a slow velocity and gradually speed up which is represented by the steep downward curve of the graph. 

Velocity-Time Graph for an Object in Free Fall

As an object falls, its velocity increases due to the acceleration of gravity. This means that the velocity starts slow and steadily increases in the downward direction. The graph below shows the velocity-time for an object in free fall:

Note that the slope of this graph is constant and represents the acceleration due to gravity, or -9.81\text{ m/s}^2 .

Acceleration-Time Graph for an Object in Free Fall

Free fall acceleration is constant. Throughout the entire time that an object is falling, it is accelerating at a rate equal to the acceleration due to gravity, -9.81\text{ m/s}^2 . As shown in the graph below, the acceleration-time graph is a constant negative line. 

Projectile Motion

A projectile is an object that is launched or thrown into the air and then only influenced by gravity. Projectile motion has many similarities to free fall motion, however, projectiles may also travel a horizontal distance in addition to falling vertically down. 

Examples of Projectile Motion

The exact trajectory, or path, a projectile will take depends on how it is launched. However, all projectiles follow a curved trajectory such as in the image shown below:

If you play or watch sports, you likely have already observed projectile motion. Projectile motion describes the arc of a basketball in a free throw, a fly ball in baseball, or a volleyball bumped over the net. 

Horizontal Component of Velocity

To analyze projectile motion, we must separate the motion into horizontal and vertical components. The horizontal component of a projectile’s velocity is independent of the vertical component of velocity. Since gravity acts vertically, there are no horizontal forces acting on projectiles. This means that the horizontal component of a projectile’s velocity remains constant throughout the entire flight. 

Example: Finding the Horizontal Component

For example, a projectile is launched from the ground with an initial speed of 8\text{ m/s} at a 60^{\circ} angle. What is the horizontal component of the projectile’s velocity?

We will need to use trig identities to determine the components of the velocity. We can visualize the components as a triangle where the hypotenuse is the initial velocity and the sides represent the horizontal, v_{ix} , and vertical, v_{iy} , components of the velocity.

Cosine is defined as the adjacent side of the triangle divided by the hypotenuse. Since the horizontal component is adjacent to the angle, we can use cosine to find the horizontal component of velocity:

Therefore, the horizontal component of the initial velocity is 4\text{ m/s} .

Need to review your trig identities? Try out this resource from Khan Academy .

Vertical Component of Velocity

The vertical component of a projectile’s velocity will be influenced by gravity, which acts vertically on the object causing it to accelerate downward. Therefore, the vertical component of velocity will change throughout the projectile’s flight. We can calculate the vertical component of velocity at a particular time in a method similar to calculating the horizontal component. 

Example: Finding the Vertical Component

In the same example, a projectile is launched from the ground with an initial speed of 8\text{ m/s} at a 60^{\circ} angle. What is the vertical component of the projectile’s velocity?

As we visualize the velocity components, we are solving this time for the opposite side of the triangle. Sine is defined as the opposite side of the triangle divided by the hypotenuse. Therefore, the initial vertical velocity is:

Solving Projectile Motion Questions

Let’s apply what we’ve learned to some examples of projectile motion!

Example 1: Finding the Range of a Projectile

In this example, a projectile is fired horizontally with a speed of 5\text{ m/s} from a cliff with a height of 60\text{ m} . How far from the base of the cliff will the projectile land? 

In this scenario, we are given the initial horizontal velocity v_{ix}=5\text{ m/s} and the vertical change in position d_y=-60\text{ m} . Since the projectile is launched horizontally, the initial vertical velocity, v_{iy} , is zero. We also always know in projectile motion that the vertical acceleration is a_y=-9.81\text{ m/s}^2 and the horizontal acceleration, a_x , is zero.

This problem is asking us to find the horizontal displacement, or d_x . This is also referred to as the range . We can use the following kinematic equation to find the projectile’s final horizontal position:

Since the horizontal acceleration of a projectile is zero, this equation can be simplified to:

Before we can solve this equation, we must first determine the time of the projectile’s flight. We can actually use this same equation in the vertical direction to solve for time:

Since the initial vertical velocity is zero, this equation can be simplified to:

Solving for t :

Substituting the given values:

Now we can use this time to calculate the horizontal displacement of the projectile:

Therefore, the projectile will land about 17.5\text{ m} from the base of the cliff. 

Example 2: Finding the Maximum Height of a Projectile

As another example, a projectile is launched from the ground with an initial velocity of 25\text{ m/s} at an angle of 50^{\circ} . What is the projectile’s maximum height?

As a projectile travels upward, its vertical velocity becomes slower and slower due to the negative acceleration of gravity. At the maximum height of the trajectory, the projectile’s vertical velocity will momentarily be zero as the projectile stops and turns to move downward. Therefore, in this scenario, our final vertical velocity, v_{fy} , is zero.

We can use the following kinematic equation to solve for the maximum height, d_y :

Solving for d_y :

Before we can use this equation to calculate the height, we will need to use the sine trig identity to find the vertical component of the initial velocity:

Since the initial velocity is in the opposite direction as the acceleration, it’s really important to remember the sign here. If we define moving up as positive, then the initial velocity is positive and the acceleration is negative. Substituting this initial vertical velocity and the given values into the equation above gives:

Therefore, the projectile will reach a maximum height of about 18.7\text{ m} .

For more examples and an explanation of solving these types of projectile motion problems, check out this youtube video from Professor Dave . 

Understanding free fall and projectile motion allows you to solve some of the most complex problems you will encounter in introductory physics. All projectiles are acted on only by gravity, and the vertical and horizontal components of motion are independent of each other. This allows us to apply our kinematic equations to solve for a projectile’s time of flight, velocity, and displacement in each direction.  

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Mechanics: 1-Dimensional Kinematics

Calculator pad, version 2, 1-d kinematics problem set.

In the 2008 Olympics, Jamaican sprinter Usain Bolt shocked the world as he ran the 100-meter dash in 9.69 seconds. Determine Usain's average speed for the race.

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In the Funny Car competition at the Joliet Speedway in Joliet, Illinois in October of 2004, John Force complete the ¼-mile dragster race in a record time of 4.437 seconds. Determine the average speed of the dragster in mi/hr and m/s. GIVEN: (1.000 mi =1609 m)

In the qualifying round of the 50-yd freestyle in the sectional swimming championship, Dugan got an early lead by finishing the first 25.00 yd in 10.01 seconds. Dugan finished the return leg (25.00 yd distance) in 10.22 seconds.

a. Determine Dugan's average speed for the entire race. b. Determine Dugan's average speed for the first 25.00 yd leg of the race. c. Determine Dugan's average velocity for the entire race.

In last week's Homecoming victory, Al Konfurance, the star halfback of South's football team, broke a tackle at the line of scrimmage and darted upfield untouched. He averaged 9.8 m/s for an 80-yard (73 m) score. Determine the time for Al to run from the line of scrimmage to the end zone.

During the annual shuffleboard competition, Renee gives her puck an initial speed of 9.32 m/s. Once leaving her stick, the puck slows down at a rate of -4.06 m/s/s.

a. Determine the time it takes the puck to slow to a stop. b. Use your initial speed and the calculated time to determine the average speed and the distance which the puck travels before stopping.

Ken Runfast is the star of the cross-country team. During a recent morning run, Ken averaged a speed of 5.8 m/s for 12.9 minutes. Ken then averaged a speed of 6.10 m/s for 7.1 minutes. Determine the total distance which Ken ran during his 20 minute jog.

The Lamborghini Murcielago can accelerate from 0 to 27.8 m/s (100 km/hr or 62.2 mi/hr) in a time of 3.40 seconds. Determine the acceleration of this car in both m/s/s and mi/hr/s.

Homer Agin leads the Varsity team in home runs. In a recent game, Homer hit a 96 mi/hr sinking curve ball head on, sending it off his bat in the exact opposite direction at 56 mi/hr. The actually contact between ball and bat lasted for 0.75 milliseconds. Determine the magnitude of the average acceleration of the ball during the contact with the bat. Express your answer in both mi/hr/s and in m/s/s. (Given: 1.00 m/s = 2.24 mi/hr)

A Formula One car is a single-seat racing car with an open cockpit and substantial wings located in the front and rear. At high speeds, the aerodynamics of the car help to create a strong downward force which allows the car to brake from 27.8 m/s (100 km/hr or 62.2 mi/hr) to 0 in as small of a distance as 17 meters. Determine the deceleration rate (i.e., acceleration) achieved by such a car.

The position-time graph below represents the motion of South's basketball coach during the last sixteen seconds of overtime during this past weekend's game.

Use the graph to answer the next several questions.

a. Determine the total distance walked by the coach during these 16 seconds. b. Determine the resulting displacement of the coach during these 16 seconds. c. Determine the displacement of the coach after 12.0 seconds. d. At what time did the coach have the greatest displacement from his starting position? e. What was the fastest speed which the coach walked during any of the time intervals for the last 16.0 seconds? f. What was the average speed of the coach for these 16.0 seconds?

Mr. H is preparing to show the class a Strobe demonstration when he realizes that his absent-mindedness has struck once more. He left the strobe on the counter in the back of the lab after the last class period. Starting 1.0 meter from the front of the room, Mr. H walks quickly to the back of the lab, picks up the strobe and returns to the middle of the classroom. The position-time graph below represents his motion. Use the graph to answer the next several questions.

a. What is the total distance walked by Mr. H during these 8.0 seconds? b. What is the average speed of Mr. H during these 8.0 seconds? c. What is the average velocity of Mr. H during these 8.0 seconds? d. How fast did Mr. H walk during the first 5.0 seconds? e. How fast did Mr. H walk during the last 3.0 seconds?

The position-time graph below represents the motion of two students - Mac (in red) and Tosh (in blue) - as they enter and exit the school library during a passing period.

Use the graph to determine the speeds at which the two students move. (Ignore any stationary periods of time.) Then determine how much faster the fastest student moves than the slower student.

Renatta Gas did it again. She failed to fill up her tank during the last four weeks. The velocity-time graph below represents the last six seconds of motion her car before being stranded on a highway in route to her university.

Use this graph to determine...

a. ...the acceleration of Renatta's car. b. ...the distance traveled during her last 6.0 seconds of motion.

Marcus Tardee is driving his friends to school. With the start of school being only minutes away, he is unfortunately following a slow garbage truck. The truck finally turns down a side street and Marcus accelerates to a much more customary speed. The velocity-time graph below represents his motion. Use the graph to answer the following questions.

a. How fast was Marcus traveling while following the garbage truck? b. Determine the distance traveled during the first 4.0 seconds represented on the graph. c. Determine the acceleration of the car once the garbage truck turned onto the side street. d. Determine the distance traveled by the car during the last 6.0 seconds of motion.

The velocity-time graph below represents the motion of a car on a city street.

Use the graph to determine the acceleration values of the car at ...

a. 1.4 seconds. b. 6.8 seconds. c. 11.6 seconds. d. 17.6 seconds.

After a long soccer practice down at the neighborhood soccer fields, Suzie begins walking up the steep hill towards her home. She gives her soccer ball a kick up the hill and continues walking towards it, meeting the ball as it is rolling back down. The velocity-time graph below depicts the motion of the ball. Use the graph to answer the following questions.

a. At what time did the ball change directions and begin rolling back down the hill? b. What is the acceleration of the ball as it rolls up the hill? down the hill? c. How far up the hill did the ball roll before it began to roll back down? d. Determine the total distance traveled by the ball during the 5.00 seconds - both the distance up the hill and down the hill. e. How far up the hill did Suzie walk between the time when she kicked the ball and the time she met up with the ball (at 5.0 seconds)?

Jeremy has recently taken up snowboarding as a hobby. He is practicing making smooth turns while traveling up sloped inclines. The velocity-time graph below depicts his motion traveling up an embankment and part-way down. Use the graph to answer the following questions.

a. Determine Jeremy's acceleration at 8.0 seconds. b. Determine the distance Jeremy traveled from 0.0 to 5.0 seconds. c. At what time did Jeremy begin to travel back down the embankment?

A Cessna 150 airplane has a takeoff speed of 28 m/s (63 mi/hr). Determine the minimum length of the runway which would be required for the plane to take off if it averages an acceleration of 1.9 m/s/s.

Cynthia competes in luge competitions during the winter months. She rides solo on a small sled 3 inches off the ground down icy slopes, turning only by use of her feet and the shifting of her weight on the sled. During the initial stage of one downhill luge, Cynthia accelerated from rest at 6.84 m/s/s for 2.39 seconds. Determine the distance she moved during this acceleration phase.

Suzie Lavtaski has reached the end of the ski slope and abruptly decelerates from 29.0 m/s to 1.8 m/s in 1.45 seconds. Determine Suzie' acceleration rate and the distance she moved during this braking period.

Captain John Stapp is often referred to as the "fastest man on Earth." In the late 1940s and early 1950s, Stapp ran the U.S. Air Force's Aero Med lab, pioneering research into the accelerations which humans could tolerate and the types of physiological effects which would result. After several runs with a 185-pound dummy named Oscar Eightball, Captain Stapp decided that tests should be conducted upon humans. Demonstrating his valor and commitment to the cause, Stapp volunteered to be the main subject of subsequent testing.

Manning the rocket sled on the famed Gee Whiz track, Stapp tested acceleration and deceleration rates in both the forward-sitting and backward-sitting positions. He would accelerate to aircraft speeds along the 1200-foot track and abruptly decelerate under the influence of a hydraulic braking system. On one of his most intense runs, his sled decelerated from 282 m/s (632 mi/hr) to a stop at -201 m/s/s. Determine the stopping distance and the stopping time.

Julietta and Jackson are playing miniature golf. Julietta's ball rolls into a long. straight upward incline with a speed of 2.95 m/s and accelerates at -0.876 m/s/s for 1.54 seconds until it reaches the top of the incline and then continues along an elevated section. Determine the length of the incline.

Rickey Henderson, baseball's record holder for stolen bases, approaches third base. He dives head first, hitting the ground at 6.75 m/s and reaching the base at 5.91 m/s, accelerating at -5.11 m/s/s. Determine the distance Rickey slides across the ground before touching the base.

Win Blonehare and Kent Swimtashore are sailboating in Lake Gustastorm. Starting from rest near the shore, they accelerate with a uniform acceleration of 0.29 m/s/s, How far are they from the shore after 18 seconds?

For years, the tallest tower in the United States was the Phoenix Shot Tower in Baltimore, Maryland. The shot tower was used from 1828 to1892 to make lead shot for pistols and rifles and molded shot for cannons and other instruments of warfare. Molten lead was dropped from the top of the 234-foot (71.3 meter) tall tower into a vat of water. During its free fall, the lead would form a perfectly spherical droplet and solidify. Determine the time of fall and the speed of a lead shot upon hitting the water at the bottom.

According to Guinness, the tallest man to have ever lived was Robert Pershing Wadlow of Alton, Illinois. He was last measured in 1940 to be 2.72 meters tall (8 feet, 11 inches). Determine the speed which a quarter would have reached before contact with the ground if dropped from rest from the top of his head.

A California Condor is approaching its nest with a large chunk of carrion in its beak. As it approaches, it makes an upward swoop, achieving a momentary upward velocity of 12.8 m/s when the carrion falls from its mouth, hitting a cliff outcropping 32.1 m below. Determine the speed of the carrion upon hitting the outcropping.

During his recent skydiving adventure, Luke Autbeloe had reached a terminal speed of 10.4 m/s as he approached the ground with his parachute. During an attempt to snap one last photo with his camera, Luke fumbled it from a height of 52.1 m above the ground.

a. Determine the speed with which the camera hits the ground. b. Determine the time for the camera to free fall from Luke's hands to the ground.

The speed required of a military jet when taking off from the deck of an aircraft carrier is dependent upon the speed of the carrier and the speed of the wind into which the carrier is moving. The takeoff speed required of a military jet relative to the deck of the carrier is 45 m/s when the carrier travels at 45 mi/hr into a 20 mi/hr wind. And when the aircraft carrier is traveling at 10 mi/hr into a 5 mi/hr wind, the takeoff speed relative to the deck of the carrier is 71 m/s. Determine the acceleration which a military jet must have to take off under these two conditions from the 126-m long runway of the USS Ronald Reagan aircraft carrier.

The Zero Gravity Research Facility at NASA-operated Glenn Research Center in Ohio is used to test the behavior of fluids, flames, equipment and other objects in free fall. It consists of a 467-foot long, 12-foot diameter, steel vacuum chamber. The steel chamber resides inside of a concrete lined shaft which extends 510 feet below ground level. Objects falling through the tower experience free fall over a distance of 432 feet (132 meters).

a. Determine the falling time for objects dropped from rest. b. Determine the final speed of the objects before the braking period begins.

It's breakfast time and Mr. H entertains himself once more by watching the daily beetle race across the 35.7-cm length of the Wheaties box top. Angie the beetle typically averages 3.77 mm/s and Bessie the beetle averages 4.78 mm/s. If Bessie gives Angie a 5.4 cm head start, then which beetle wins and by what distance?

Alexander's hobby is dirt biking. On one occasion last weekend, he accelerated from rest to 17.8 m/s/s in 1.56 seconds. He then maintained this speed for 9.47 seconds. Seeing a coyote cross the trail ahead of him, he abruptly stops in 2.79 seconds. Determine Alexander's average speed for this motion.

Ima Rushin can travel from Milwaukee Avenue to the school entrance gate at a constant speed of 22.5 m/s when the lights are green and there is no traffic. On Wednesday, Ima is stopped by a red light at Landwehr Road. She decelerates at -3.95 m/s/s, waits for 45.0 seconds before the light turns green and accelerates back up to speed at 4.91 m/s/s.

a. Determine the total time required to decelerate, stop and accelerate back up to speed. b. Determine the amount of time the red light costs the driver (compared to if the car had not been stopped by the red light).

A tortoise and a hare are having a 1000-meter race. The tortoise runs the race at a constant speed of 2.30 cm/s. The hare moves at an average speed of 1.50 m/s for 10.0 minutes and then decides to take a nap. After waking up from the nap, the hare recognizes that the tortoise is about to cross the finish line and immediately accelerates from rest with a constant acceleration of 0.500 m/s/s for the remaining distance of the race. If the tortoise wins by a hair (no pun intended), then what is the time in hours that the hare napped?

Hayden and Matthew are riding around the neighborhood on their scooters. Hayden is at rest when Matthew passes him moving at a constant speed of 0.37 m/s. After 1.8 seconds, Hayden decides to chase after Matthew, accelerating at 0.91 m/s/s. How much time must Hayden accelerate before he is side-by-side with Matthew?

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