Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 3 - Kinematics in Two Dimensions; Vectors - Problems: 34

Answer

$t_{up} = V_{y0}$$/g ; t_{down} = V_{y0}$$/g$

Work Step by Step

Choose x = 0 and y = 0 to be the point of launch and upward to be the positive y direction. Let the initial vertical velocity be $V_{y0}$. Use equation 2.11a to find the time to reach the highest point, at which the vertical velocity is zero. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. The time moving up is $V_{y0}$$/g$. Now use equation 2.11c to find the maximum height reached: this is $\frac{Vy0^{2}}{2g}$. Finally, use equation 2.11b to find the time for the object to fall that distance with a starting vertical velocity of 0. The time to fall is $V_{y0}$$/g$, which was to be proven.
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