Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 6 - Work and Energy - Problems - Page 166: 54

Answer

The speed at the bottom of the cliff is 173 m/s.

Work Step by Step

The kinetic energy at the bottom will be equal to the kinetic energy plus the potential energy at the top. Therefore, $KE_2 = KE_1 + PE$ $\frac{1}{2}mv_2^2 = \frac{1}{2}mv_1^2 + mgh$ $v_2 = \sqrt{v_1^2 + 2gh}$ $v_2 = \sqrt{(165 ~m/s)^2 + (2)(9.8 ~m/s^2)(135 ~m)}$ $v_2 = 173 ~m/s$ The speed at the bottom of the cliff is 173 m/s.
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