Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems: 16

Answer

(a) t = 15.3 seconds (b) $x = 330~m$ (c) The ball would travel six times farther on the moon.

Work Step by Step

(a) We can find the acceleration of gravity on the moon. $g_m = \frac{9.80~m/s^2}{6} = 1.63~m/s^2$ We can find $t_{up}$, the time for the ball to reach maximum height. $t_{up} = \frac{v_y-v_{y0}}{g_m}$ $t_{up} = \frac{0-(25~m/s)~sin(30^{\circ})}{-1.63~m/s^2}$ $t_{up} = 7.67~s$ The total time is $2~t_{up}$ which is 15.3 seconds. (b) We can find the distance $x$ the ball travels on the moon. $x = \frac{v_0^2~sin(2\theta)}{(g/6)}$ $x = \frac{(6)(25~m/s)^2~sin(60^{\circ})}{9.80~m/s^2}$ $x = 330~m$ (c) We can find the distance the golf ball travels on earth. $x = \frac{v_0^2~sin(2\theta)}{g}$ $x = \frac{(25~m/s)^2~sin(60^{\circ})}{9.80~m/s^2}$ $x = 55~m$ The ball would travel six times farther on the moon.
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