Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson

Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems - Page 112: 43

Answer

(a) The flight time is 15.3 seconds. (b) $x = 331~m$ (c) The ball would travel 276 meters 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. $t_{up} = \frac{v_{y0}}{a}$ $t_{up} = \frac{(25~m/s)~sin(30^{\circ})}{1.63~m/s^2}$ $t_{up} = 7.67~s$ The total time is $2~t_{up}$ which is 15.3 seconds. (b) $x = v_x~t$ $x = (25~m/s)~cos(30^{\circ})(15.3~s)$ $x = 331~m$ (c) We can find the range of the golf ball on earth. $range = \frac{v_0^2~sin(2\theta)}{g}$ $range = \frac{(25~m/s)^2~sin(60^{\circ})}{9.80~m/s^2}$ $range = 55~m$ The ball would travel 276 meters farther on the moon.

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