   Chapter 3, Problem 34AP

Chapter
Section
Textbook Problem

You can use any coordinate system you like to solve a projectile motion problem. To demonstrate the truth of this statement, consider a ball thrown off the top of a building with a velocity v → at an angle θ with respect to the horizontal. Let the building be 50.0 m tall, the initial horizontal velocity be 9.00 m/s, and the initial vertical velocity be 12.0 m/s. Choose your coordinates such that the positive y-axis is upward, the x-axis is to the right, and the origin is at the point where the ball is released, (a) With these choices, find the ball’s maximum height above the ground and the time it takes to reach the maximum height. (b) Repeat your calculations choosing the origin at the base of the building.

(a)

To determine
The ball’s maximum height above the ground and the time it takes to reach the maximum height above the ground and the time taken to reach the maximum height.

Explanation

The vertical displacement from the launch point (top of the building) to the top of the arc is,

Δy=vy2v0y22ay

Here,

vy is the vertical component of the velocity

v0y is the vertical component of the initial velocity

ay is the vertical acceleration

Substitute 0m/s for vy , 12.0m/s for v0y and 9.8m/s2 for ay .

Δy=(0m/s)2(12.0m/s)22(9.80m/s2)=7.35m

The following diagram shows the projection of ball from the top of the building.

From the figure, the maximum height is,

hmax=50

(b)

To determine
The ball’s maximum height above the ground and the time it takes to reach the maximum height above the ground and the time taken to reach the maximum height choosing the origin at the base of the building.

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