Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 4 - Motion in Two and Three Dimensions - Problems - Page 86: 42b



Work Step by Step

At any given time, the height of the particle will be: $H_{t}=H_{0}+v_{0}\sin \theta _{0}t-\dfrac {gt^{2}}{2}\left.....( 1\right) $ where $H_{0}$ is initial height of particle, $v_{0}$ is initial velocity, $\theta_{0}$ is the initial angle of particle relative to horizontal line and $t$ is the time passed since the object launched. Lets calculate the maximum height of the object: $H_{\max }=H_{0}+\dfrac {v^{2}_{0}\sin ^{2}\theta _{0}}{2g}=3m+\dfrac {\left( 26.5\dfrac {m}{s}\right) ^{2}\sin ^{2}53^{\circ}}{2\times 9.8\dfrac {m}{s^{2}}}=25.85m$ Therefore, $\Delta H=H_{\max -}H_{wheel}=25.85-18=7.85m$
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.