Physics: Principles with Applications (7th Edition)

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

Chapter 3 - Kinematics in Two Dimensions; Vectors - General Problems: 64

Answer

The minimum pushoff speed is 1.9 m/s. They are in the air for 2.7 seconds.

Work Step by Step

We can find an expression for the time it takes to fall a vertical distance $y$ using the formula: $y = \frac{1}{2}gt^2$ $t = \sqrt{\frac{2y}{g}}$ We can then use the time t to find the minimum pushoff speed; $v_x = \frac{x}{t}$ $v_x = x~\sqrt{\frac{g}{2y}}$ $v_x = (5.0~m)~\sqrt{\frac{9.80~m/s^2}{(2)(35~m)}}$ $v_x = 1.9~m/s$ We can find the time that they are in the air: $t = \sqrt{\frac{2y}{g}}$ $t = \sqrt{\frac{(2)(35 ~)}{9.80 ~m/s^2}} = 2.7 ~s$
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