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Chapter 7: Problem 66

A golf ball is released from rest from a height of \(0.811 \mathrm{~m}\) above the ground and has a collision with the ground, for which the coefficient of restitution is \(0.601 .\) What is the maximum height reached by this ball as it bounces back up after this collision?

Short Answer

Expert verified
Answer: The maximum height reached by the golf ball after the bounce is 0.293 meters.

Step by step solution

01

Find the initial velocity of the ball before impact

We'll use the following kinematic equation to find the initial velocity of the ball just before it hits the ground, with \(h = 0.811 \mathrm{~m}\): \(v^2 = u^2 + 2as\), where \(v\) is the final velocity, \(u\) is the initial velocity (0 m/s as the ball is released from rest), \(a\) is the acceleration (9.81 m/s², due to gravity), and \(s\) is the height. Solving for \(v\), we get: \(v= \sqrt{u^2+2as} = \sqrt{(0)^2+2(9.81)(0.811)}\)
02

Calculate the initial velocity before impact

Using the previous equation, we find the initial velocity of the ball just before hitting the ground: \(v=\sqrt{(0)^2+2(9.81)(0.811)}=4 \mathrm{m/s}\)
03

Use the coefficient of restitution to find the velocity after collision

The coefficient of restitution (\(e\)) is given by \(0.601\). The formula relating the relative velocities of two objects before (V1, V2) and after (V1', V2') a collision is: \(e=\frac{V2'-V1'}{V1-V2}\) Since the ground is not moving, both V2 and V2' are 0. Therefore: \(e = \frac{-V1'}{V1}\) Solving for \(V1'\) (the golf ball's velocity after the collision), we get: \(V1'=-e \cdot V1= -0.601 \cdot 4\mathrm {m/s} = -2.404 \mathrm{m/s}\) Note that the velocity is negative since it is in the opposite direction of the initial velocity.
04

Calculate the maximum height reached after the bounce

Now we will use the kinematic equation again, but this time we will set the final velocity (\(v\)) to 0 m/s (because at the maximum height, the velocity is 0). Our initial velocity is now the velocity after the collision (\(V1'\)) and we are solving for \(s\), the height: \(0 = (-2.404)^2 + 2(-9.81)s\) Rearrange the equation to solve for s: \(s= \frac{(-2.404)^2}{2(9.81)}\)
05

Calculate the maximum height reached

Using the previous equation, we find the maximum height reached by the golf ball after it bounces: \(s=\frac{(-2.404)^2}{2(9.81)}= 0.293 \mathrm{~m}\) Thus, the maximum height reached by the golf ball after the bounce is 0.293 meters.

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