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
$x=v_{\circ}t$
$\implies x=v_{\circ}\sqrt{\frac{2h}{g}}$
$x=\sqrt{\frac{2v_{\circ}^2 h}{g}}$
Essential University Physics: Volume 1 (3rd Edition)
by Wolfson, Richard
Published by
Pearson
ISBN 10:
0321993721
ISBN 13:
978-0-32199-372-4
Chapter 3 - Exercises and Problems - Page 48: 64
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