Physics: Principles with Applications (7th Edition)

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

Chapter 2 - Describing Motion: Kinematics in One Dimension - General Problems - Page 48: 80

Answer

(a) The second child reaches a maximum height of 0.82 meters. (b) The first child's initial speed is 4.8 m/s. (c) The first child's total time in the air is 0.98 seconds.

Work Step by Step

(a) $y = \frac{v^2-v_0^2}{2a} = \frac{0-(4.0 ~m/s)^2}{(2)(-9.80 ~m/s^2)} = 0.82 ~m$ The second child reaches a maximum height of 0.82 meters. (b) The first child reaches a maximum height that is 1.5 times the second child's maximum height. $y = (1.5)(0.82 ~m) = 1.2 ~m$ We can use the first child's maximum height to find the first child's initial speed. $v_0^2 = v^2 - 2ay = 0 -2ay = -2ay$ $v_0 = \sqrt{-2ay} = \sqrt{(-2)(-9.80 ~m/s^2)(1.2 ~m)} = 4.8 ~m/s$ The first child's initial speed is 4.8 m/s. (c) We can use the initial speed to find the time it takes to reach the maximum height. $t = \frac{v-v_0}{a} = \frac{0 -4.8 ~m/s}{-9.80 ~m/s^2} = 0.49 ~s$ The total time in the air is 2t which is 0.98 seconds.
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