College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 4 - Problems - Page 151: 56

Answer

(a) At the highest point, $v_x = v_i~cos~\theta$ At the highest point, $v_y = 0$ (b) $t = \frac{v_i~sin~\theta}{g}$ (c) $H = \frac{(v_i~sin~\theta)^2}{2g}$

Work Step by Step

(a) Since $v_x$ is constant, at the highest point, $v_x = v_i~cos~\theta$ At the highest point, $v_y = 0$ (b) At the maximum height, $v_y = 0$. We can find the time to reach the maximum height: $v_y = v_{0y} +at$ $t = \frac{v_y-v_{0y}}{a}$ $t = \frac{0-v_i~sin~\theta}{-g}$ $t = \frac{v_i~sin~\theta}{g}$ (c) We can find he maximum height $H$: $v_y^2 = v_{0y}^2+2a_yH$ $H = \frac{v_y^2-v_{0y}^2}{2a_y}$ $H = \frac{0-(v_i~sin~\theta)^2}{(2)(-g)}$ $H = \frac{(v_i~sin~\theta)^2}{2g}$
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.