College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 3 - Motion in Two Dimensions - Learning Path Questions and Exercises - Exercises - Page 100: 59



Work Step by Step

Let angle of initial velocity be $\theta$ with respect to the horizontal. $v^{2}=u^{2}+2aH$ $0=(usin\theta)^{2}-2gH$ $H=\frac{u^{2}sin\theta^{2}}{2g}$ Now, $v=u+at$ $0=usin\theta-gt$ $t=\frac{usin\theta}{g}$ So, $R=ucos\theta \times t=ucos\theta \times \frac{usin\theta}{g}$ Therefore, $H=\frac{1}{2} R$ or, $\frac{u^{2}sin\theta^{2}}{2g}=\frac{1}{2}ucos\theta \times \frac{usin\theta}{g}$ or, $tan\theta=1$ So, $\theta=45^{\circ}$
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.