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

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 2 - Kinematics in One Dimension - Exercises and Problems - Page 64: 17

Answer

(a) The tower should be 78.4 meters high. (b) The impact velocity is 39.2 m/s

Work Step by Step

(a) $y = \frac{1}{2}at^2$ $y = \frac{1}{2}(9.80~m/s^2)(4.0~s)^2$ $y = 78.4~m$ The tower should be 78.4 meters high. (b) $v = v_0+at = 0 + at$ $v = (9.80~m/s^2)(4.0~s)$ $v = 39.2~m/s$ The impact velocity is 39.2 m/s
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