# 3.4: Projectile Motion

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 illustrates the notation for displacement, where *s*, *x*, and *y*. (Note that in the last section we used the notation

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:

The total displacement

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

*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

*Step 4.* *Recombine the two motions to find the total displacement* * and velocity **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

Total displacement and velocity

(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 *x* - and *y* -motions are recombined to give the total velocity at any given point on the trajectory.

Example 3.5.1: 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

Strategy

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

Solution for (a)

By “height” we mean the altitude or vertical position

<figcaption>

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

Solving for

Now we must find *y*-direction. It is given by

and

so that

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

Note that the final vertical velocity,

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

Solution for (c)

Because air resistance is negligible,

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

The time

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

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

Exercise

A common method for preparing oxygen is the decompositio

It is important to set up a coordinate system when analyzing projectile motion. One part of defining the coordinate system is to define an origin for the

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

<figcaption>

The trajectory of a rock ejected from the Kilauea volcano.

</figcaption> </figure>Strategy

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

Solution for (a)

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

If we take the initial position

Rearranging terms gives a quadratic equation in

This expression is a quadratic equation of the form

This equation yields two solutions:

Discussion for (a)

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.

Solution for (b)

From the information now in hand, we can find the final horizontal and vertical velocities

The final vertical velocity is given by the following equation:

where

so that

To find the magnitude of the final velocity

which gives

The direction

so that

Thus,

Discussion for (b)

The negative angle means that the velocity is

One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile. On level ground, we define range to be the horizontal distance

<figcaption>

Trajectories of projectiles on level ground. (a) The greater the initial speed

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed *level ground* for which air resistance is negligible is given by

where

When we speak of the range of a projectile on level ground, we assume that

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. InAddition 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 class="ui-has-child-figcaption" id="import-auto-id1645881" style="width: 840px;"><figcaption>

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.

</figcaption> </figure>Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. Make a game out of this simulation by trying to hit a target.

<figure class="ui-has-child-figcaption" id="eip-id1462984" style="width: 660px;"> <figcaption> </figcaption> </figure># 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:
- 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 quantitiesx andy , and the components of the velocityv are given byvx=vcosθ andvy=vsinθ , wherev is the magnitude of the velocity andθ is its direction. - Analyze the motion of the projectile in the horizontal direction using the following equations:
Horizontal motion(ax=0) x=x0+vxt vx=v0x=vx=velocity is a constant. - Analyze the motion of the projectile in the vertical direction using the following equations:
Vertical motion(Assuming positive direction is up;ay=−g=−9.80 m/s2) y=y0+12(v0y+vy)t vy=v0y−gt y=y0+v0yt−12gt2 v2y=v20y−2g(y−y0). - Recombine the horizontal and vertical components of location and/or velocity using the following equations:
s=x2+y2−−−−−−√ θ=tan−1(y/x) v=v2x+v2y−−−−−−√ θv=tan−1(vy/vx).

- Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position
- The maximum height
h of a projectile launched with initial vertical velocityv0y is given byh=v20y2g. - The maximum horizontal distance traveled by a projectile is called the range. The range
R of a projectile on level ground launched at an angleθ0 above the horizontal with initial speedv0 is given byR=v20sin2θ0g.

# Conceptual Questions

Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither

Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither

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?

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

A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of

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?

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?

(a) 3.50 s

(b) 28.6 m/s (c) 34.3 m/s

(d) 44.7 m/s,

(a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a

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?

(a)

(b) The arrow will go over the branch.

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?

Verify the ranges for the projectiles in Figure(a) for

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

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

(a) 560 m/s

(b)

(c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b).

An arrow is shot from a height of 1.5 m toward a cliff of height

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,

1.50 m, assuming launch angle of

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.

Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle

yes, the ball lands at 5.3 m from the net

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

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.

(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.

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.

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

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

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

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.

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

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.

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

15.0 m/s

A basketball player is running at

A football player punts the ball at a

(a) 24.2 m/s

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

Prove that the trajectory of a projectile is parabolic, having the form

Derive

so that

since

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.

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.

## Glossary

- air resistance
- a frictional force that slows the motion of objects as they travel through the air; when solving basic physics problems, air resistance is assumed to be zero

- kinematics
- the study of motion without regard to mass or force

- motion
- displacement of an object as a function of time

- projectile
- an object that travels through the air and experiences only acceleration due to gravity

- projectile motion
- the motion of an object that is subject only to the acceleration of gravity

- range
- the maximum horizontal distance that a projectile travels

- trajectory
- the path of a projectile through the air

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