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 - Problems - Page 45: 44

Answer

The initial launch velocity is 4.85 m/s and the time in the air is 0.99 seconds.

Work Step by Step

(a) $v_0^2 = 2ay$ $v_0 = \sqrt{2ay} = \sqrt{(2)(9.80)(1.20)} = 4.85~m/s$ (b) The time $t$ it takes to reach maximum height is: $t = \frac{v_0}{g} = \frac{4.85}{9.80} = 0.495 ~s$ The total time in the air is $2t$ which is 0.99 seconds.
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