Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 15 - Kinetics of a Particle: Impulse and Momentum - Section 15.3 - Conservation of Linear Momentum for a System of Particles - Problems - Page 263: 45


$v_2=\sqrt{v_1^2+2gh}$ $ \theta_2=sin^{-1}(\frac{v_1 sin\theta_1}{\sqrt{v_1^2+2gh}})$

Work Step by Step

The required speed and angle can be determined as follows: According to the law of conservation of energy $\frac{1}{2}mv_1^2+mgh=\frac{1}{2}mv_2^2+0$ This simplifies to: $v_2=\sqrt{v_1^2+2gh}$ Now, according to the principle of impulse and momentum $mv_1sin\theta_1=mv_2sin\theta_2$ $\implies sin\theta_2=\frac{v_1sin\theta_1}{\sqrt{v_1^2+2gh}}$ $\implies \theta_2=sin^{-1}(\frac{v_1 sin\theta_1}{\sqrt{v_1^2+2gh}})$
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.