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 - Page 49: 68


The proof is below.

Work Step by Step

We know the following equation: $2gh = v_0^2 sin^2\theta$ We do not need the $v_0^2$ term, so we want to find a way to cancel it out. Thus, we use the following equation: $x = \frac{v_0^2 sin(2\theta)}{g}$ Solving for $v_0^2$ gives: $v_0^2 = \frac{gx}{sin(2\theta)}$ We substitute this into the first equation to obtain: $2gh = \frac{gx}{sin(2\theta)} sin^2\theta$ $2h = \frac{x}{sin(2\theta)} sin^2\theta$ $\frac{2hsin(2\theta)}{sin^2\theta} = x$ $\frac{4hsin(\theta)cos(\theta)}{sin^2\theta} = x$ $x=\frac{4h}{tan\theta}$
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