   Chapter 4, Problem 15P

Chapter
Section
Textbook Problem

After falling from rest from a height of 30.0 m, a 0.500-kg ball rebounds upward, reaching a height of 20.0 m. If the contact between ball and ground lasted 2.00 ms, what average force was exerted on the ball?

To determine
The average force on the ball.

Explanation

Section 1:

To determine: The velocity of the ball when it hits the ground and when it rebounds.

The velocity of the ball when it hits the ground is 24.25 m/s.

The velocity of the ball when it rebounds is 19.8 m/s.

Explanation:

Given info: The ball is thrown from a height of 30.0 m. The mass of the ball is 0.500 kg. The rebound height is 20.0 m. The contact time between the ball and the ground is 2 ms.

Since the ball falls from rest, its initial velocity is zero.

From Newton’s equations of motion, the formula for velocity of the ball when it hits the ground is,

v1=2gy                                                       (I)

• g is the acceleration due to gravity.
• y is the vertical displacement.

The sign of g is negative because the ball is falling downwards.

Substitute 9.8ms2 forg and 30.0 m for y in Equation (I) to get v1 .

v1=2(9.8ms2)(30.0m)=24.25m/s

Thus, the velocity of the ball when it hits the ground is 24.25 m/s.

From Newton’s equations of motion, the formula for velocity of the ball when it rebounds is,

v2=2gyr                                                       (II)

• g is the acceleration due to gravity.
• yr is the rebound height.

The sign of g is negative because the ball is falling downwards.

Substitute 9.8ms2 for g and 20.0 m for y in Equation (II) to get v2 .

v2=2(9.8ms2)(20.0m)=19.8m/s

Thus, the velocity of the ball when it rebounds is 19.8 m/s.

Section 2:

To determine: The average acceleration of the ball.

Answer: The average acceleration of the ball is 2.2×104ms2 .

Explanation:

The formula to calculate the average acceleration of the ball is,

aavg=ΔvΔt

• Δv is the change in velocity which is equal to v2v1

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