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 - Page 70: 29



Work Step by Step

Let x= 0 and y = 0 where the shot put is launched, and let upward be the positive y direction. The projectile’s initial vertical velocity is $(14.4 m/s)(sin 34 ^{\circ})$. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. The final vertical position is -2.1 m. Use equation 2.11b to find the time of flight of 1.87 s. The other root, a negative time, is discarded. The horizontal distance covered is the constant horizontal velocity multiplied by the time. $$(14.4 m/s)(cos 34 ^{\circ})(1.87 s) = 22.3 m.$$
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