Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 7 - Exercises and Problems - Page 131: 43

Answer

Please see the work below.

Work Step by Step

We know that $K_U=K_{\circ}+U_{\circ}$ $\implies (\frac{1}{2})mv^2+0.0=(\frac{1}{2})mv_{\circ}^2+mgh$ $\frac{1}{2}mv^2=\frac{1}{2}mv_{\circ}^2+mgh$ $\frac{1}{2}v^2=\frac{1}{2}v_{\circ}^2+gh$ $v^2=v_{\circ}^2+2gh$ $v=\sqrt{v_{\circ}^2+2gh}$
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