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: 77

Answer

Please see the work below.

Work Step by Step

As we know that $v_{\circ}=\sqrt{2gh}$ The maximum range is given as $R=\frac{v_{\circ}^2}{g}$ $R=\frac{(\sqrt{2gh})^2}{g}=2h$
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