University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 2 - Motion Along a Straight Line - Problems - Exercises - Page 64: 2.82


(a) The two balls collide when the time is $t = \frac{H}{v_0}$. (b) $H = \frac{v_0^2}{g}$

Work Step by Step

(a) For the first ball, $y = v_0t - \frac{1}{2}gt^2$ For the second ball, $y = H - \frac{1}{2}gt^2$ We can equate the two equations for $y$ to find where the two balls meet. $v_0t - \frac{1}{2}gt^2 = H - \frac{1}{2}gt^2$ $v_0t = H$ $t = \frac{H}{v_0}$ The two balls collide when the time is $t = \frac{H}{v_0}$ (b) $y_{max} = \frac{v_0^2}{2g}$ When $t = \frac{H}{v_0},$ $y = H - \frac{1}{2}(g)(\frac{H}{v_0})^2$ We can equate the equations for $y$ and $y_{max}$ to find the value of $H$ in terms of $v_0$ and $g$. $H - \frac{1}{2}(g)(\frac{H}{v_0})^2=\frac{v_0^2}{2g}$ Let's multiply both sides by $\frac{2v_0^2}{g}$. $\frac{2Hv_0^2}{g} - H^2=\frac{v_0^4}{g^2}$ $H^2 - \frac{2Hv_0^2}{g} +\frac{v_0^4}{g^2} = 0$ $(H - \frac{v_0^2}{g})^2 = 0$ $H = \frac{v_0^2}{g}$
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