## Physics (10th Edition)

Published by Wiley

# Chapter 7 - Impulse and Momentum - Problems - Page 192: 14

#### Answer

The height the ball was dropped is $1.36m$

#### Work Step by Step

The speed of the ball just before it hits the floor is: $$v=\frac{p}{m}=\frac{3.1}{0.6}=5.17m/s$$ We apply an equation of kinematics to find the height from which the ball was dropped. We have the ball's initial velocity $v_0=0$ and final velocity $v_f=5.17m/s$; $g=9.8m/s^2$. So the height from which the ball was dropped is $$s=\frac{v_f^2-v_0^2}{2g}=1.36m$$

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