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4-2 - 4-4 Projectile Motion

The main application that we will consider in this chapter is that of projectile motion. This motion is that of an object that is projected with some initial velocity and then allowed to free fall. Effects such as air resistance and Earth's rotation, which sometimes causes an object's motion to differ significantly from that of pure free fall, are ignored. The key point for understanding projectile motion is that the acceleration due to gravity only acts vertically and since gravity is the only influence there is no acceleration in the horizontal direction. This fact means that while there is gravitational acceleration vertically, the horizontal motion is that of constant velocity.

It is convenient to adopt a standard coordinate system for projectile motion. This standard takes the positive direction for the vertical motion upward, making downward the negative direction. This means that the acceleration of gravity will be given a negative value. For the horizontal motion, the positive direction is to the right and the negative direction to the left. Given this coordinate system we can rewrite the equations for two-dimensional motion in a form specific to projectile motion. To accomplish this rewriting we make use of several facts:

a_x = 0, v_x = v_0x = v₀ cosq,
a_y = -g, v_0y = v₀ sinq

where q is the launch angle, which is the angle of the initial velocity as measured from the horizontal direction. With these substitutions, the equations of projectile motion are

Horizontal Motion	Vertical Motion
vx = v0 cosq	vy = v0 sinq - g t
x = x0 + (v0 cosq)t

This set of equations describes projectile motion for both zero and nonzero launch angles.

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Example 4.3 Over the Edge: Wagging his tail, a dog knocks an object horizontally off the edge of a table that is 0.750 m high. If the object hits the floor 0.995 m away from the edge of the table, with what speed did it leave the table?

Picture the Problem The picture shows the object leaving the edge of the table and the path it takes. The positive coordinate directions are also indicated.

Strategy Let's choose (x₀, y₀) = (0, 0). Also notice that v_0y = 0 and cosq = 1. The only equation involving v₀ is the one x as a function of time.

Solution

1. Solve the horizontal equation for v0:
2. Use the vertical equation with all known quantities to get an expression for t.
3. Put these results together to get an expression for v0 in terms of known quantities and solve.

Insight This was an example of a projectile with horizontal launch. Here no values were calculated at an intermediate step. Instead, the formula was carried through to the end. This approach is better for handling round off error; but, sometimes problems can get messy this way.

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Example 4.4 Toss Me that Wrench: A person stands 12.5 m from the base of a building that is 40.0 m tall. The person wants to toss a wrench to his co-worker who is working on the roof. If he releases the tool at a height of 1.00 m above the ground, what velocity must he give the tool if it is to just make it onto the roof?

Picture the Problem Our sketch shows the wrench being tossed so that it just barely makes it onto the roof.

Strategy Since the wrench just barely makes it onto the roof, the top of the roof is the maximum height of the wrench where v_y = 0. Treating the motions separately, we need to find the components of v₀. From them we can determine the magnitude and direction.

Solution
Since the origin is chosen at the point of launch we take x₀ = 0, y₀ = 0, and y = 39.0 m.

1. Choose an equation containing v0y with all other quantities known and solve:
2. The only equation involving v0x requires the time. Choose a vertical equation involving time with all other quantities known:
3. Solve the horizontal equation for v0x:
4. Get the magnitude of the initial velocity:
5. Get the direction of the initial velocity:

Physlet Illustration: Projectile Motion On The Earth
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The blue ball is thrown into the air and follows a parabolic path.  How does its velocity change with time?  What is its acceleration?
Hints Is there any acceleration in the x direction? Is there any acceleration in the y direction? From the graphs, determine the acceleration.

Insight In both this example and in the previous one, the origin was taken to be the initial positions. While this is not necessary, it is sometimes useful to just adopt a convention such as this and not have to worry about what choices to make for every problem.

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Physlet Illustration: Introduction to Projectile Motion
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The blue ball is thrown into the air and follows a parabolic path.  How does its velocity change with time? Start
Hints Using techniques developed earlier, measure the x and y components of the ball's velocity near the beginning of the motion. Measure the x and y components of the ball's velocity near the end of the motion. Did the x component change? The y component? What causes the change in velocity?

Practice Quiz

A stone is thrown horizontally from the top of a 50.0 m building with a speed of 12.3 m/s. How far from the building will it land? 30.8 m 39.3 m 3.19 m 615 m

A projectile is launched from the ground with a speed of 23.7 m/s at 33.0o above the horizontal. What is its speed when it is at its maximum height above the ground? 19.9 m/s 0.00 m/s 12.9 m/s 23.7 m/s

A projectile is launched from the ground with a speed of 53.4 m/s at 68.0o above the horizontal. What is its speed when it is at its maximum height above the ground? (440 m, 440 m) (440 m, 107 m) (408 m, 333 m) (165 m, 74.9 m)

Physlet Illustration: The Monkey and the Hunter
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V = m/s q = °
A hunter walking through the forest spots a monkey in a tree.  He knows that, when he shoots, the monkey will hear the sound and immediately drop to the ground in an attempt to escape.  Where should the hunter aim?  Below the monkey?  Above the monkey? Start
Hints If one object is dropped and a second launched horizontally from the same height, which hits the ground first? How far does the monkey fall in a given time? How far does the bullet's path deviate ("fall") from a straight line in the same time period?

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