Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 4 - Kinematics in Two Dimensions - Exercises and Problems - Page 113: 45


The ball clears the net by 1.0 meter.

Work Step by Step

We can find the time for the ball to travel a horizontal distance of 7.0 meters. $t = \frac{d}{v_x} = \frac{7.0~m}{(20~m/s)~cos(5.0^{\circ})}$ $t = 0.351~s$ We can find the vertical position at this time. $y = y_0+v_{0y}t+\frac{1}{2}at^2$ $y = 2.0~m+(20~m/s)~sin(5.0^{\circ})(0.351~s)-\frac{1}{2}(9.80~m/s^2)(0.351~s)^2$ $y = 2.0~m$ Since the net is 1.0 meter high, the ball clears the net by 1.0 meter.
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.