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: 41

Answer

(a) The projectile hits the ground 81 meters lower than the launch point. (b) The projectile rises to a maximum height of 34 meters above the launch point.

Work Step by Step

(a) $y = v_{0y}t+\frac{1}{2}a_yt^2$ $y = (30~m/s)~sin(60^{\circ})(7.5~s)-\frac{1}{2}(9.80~m/s^2)(7.5~s)^2$ $y = -81~m$ The projectile hits the ground 81 meters lower than the launch point. (b) We can find the maximum height above the ground. $y = \frac{v_y^2-v_{0y}^2}{2g}$ $y = \frac{0-[(30~m/s)~sin(60^{\circ})]^2}{(2)(-9.80~m/s^2)}$ $y = 34~m$ The projectile rises to a maximum height of 34 meters above the launch point.
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.