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

Answer

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.