Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 3 - Exercises and Problems: 64

Answer

Please see the work below.

Work Step by Step

We know that $t=\sqrt{\frac{2h}{g}}$ The horizontal distance covered by the projectile is given as $x=v_{\circ}t$ $\implies x=v_{\circ}\sqrt{\frac{2h}{g}}$ $x=\sqrt{\frac{2v_{\circ}^2 h}{g}}$
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