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: 30



Work Step by Step

Let x= 0 and y = 0 directly below where the baseball is launched, and let upward be the positive y direction. The projectile’s initial vertical velocity is $(27.0 m/s)(sin 45 ^{\circ})$. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. It starts at a height of y = 1.0 m, and the final vertical position is 13.0 m. Use equation 2.11b to find the time of flight of 3.108 s. The other root, a smaller one, is discarded. It corresponds to when the baseball reached y = 13.0 m on the way up, but the baseball landed on the roof on the way down. The horizontal distance covered is the constant horizontal velocity multiplied by the time. $$(27.0 m/s)(cos 45 ^{\circ})(3.108 s) = 59.3 m.$$
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