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: 55


0.88s; 0.95m

Work Step by Step

Let x= 0 and y = 0 where the jumper starts, and let upward be the positive y direction. The vertical acceleration is $-9.8 \frac{m}{s^{2}}$, negative because it is directed downward. The time of flight is found from the constant horizontal velocity. t = 8.0 m/(9.1 m/s) = 0.88 s The time to fall from the highest height is half of that, or 0.44 s. Use equation 2.11b, with a starting vertical velocity of zero, to find a height of 0.95 m.
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