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 2 - Kinematics in One Dimension - Exercises and Problems: 29

Answer

(a) The skier's velocity at the bottom of the hill is 15.9 m/s (b) The skier can travel 30.5 meters up the slope.

Work Step by Step

(a) The acceleration directed down the incline is $g~sin(\theta)$. Therefore; $v^2 = v_0^2+2ad = 0 + 2ad$ $v = \sqrt{2ad} = \sqrt{(2)(9.80~m/s^2)~sin(15^{\circ})(50~m)}$ $v = 15.9~m/s$ The skier's velocity at the bottom of the hill is 15.9 m/s (b) The acceleration directed down the slope is $g~sin(\theta)$. So; $d = \frac{v^2-v_0^2}{2a}$ $d = \frac{0-(15.9~m/s)^2}{(2)(-9.80~m/s^2)~sin(25^{\circ})}$ $d = 30.5~m$ The skier can travel 30.5 meters up the slope.
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