## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

# Chapter 2 - Kinematics in One Dimension - Exercises and Problems - Page 68: 82

#### Answer

(a) $y = h - \frac{gh^2}{2v_0^2}$ (b) $h=\frac{2v_0^2}{g}$ (c) $h=\frac{v_0^2}{g}$

#### Work Step by Step

(a) We can write an expression for the vertical position of each ball and equate the expressions to find the time that they collide; $h-\frac{1}{2}gt^2=v_0t-\frac{1}{2}gt^2$ $t = \frac{h}{v_0}$ We can find the vertical position at this time. $y = h - \frac{1}{2}gt^2$ $y = h - \frac{1}{2}g(\frac{h}{v_0})^2$ $y = h - \frac{gh^2}{2v_0^2}$ (b) We can find $h$ when the vertical position of the collision:$y = 0$ $y = h - \frac{gh^2}{2v_0^2} = 0$ $\frac{gh^2}{2v_0^2} =h$ $h = \frac{2v_0^2}{g}$ (c) The time for the first ball to reach its highest point is $t = \frac{v_0}{g}$. We can write an expression for the vertical position of each ball at this time to find $h$; $h-\frac{1}{2}gt^2=v_0t-\frac{1}{2}gt^2$ $h=v_0(\frac{v_0}{g})$ $h=\frac{v_0^2}{g}$

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