Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Review Exercises - Page 250: 71a


$D = \frac{v^2~sin~(2~\theta)}{32}$

Work Step by Step

$D = \frac{v^2~sin~\theta~cos~\theta+v~cos~\theta~\sqrt{(v~sin~\theta)^2+64~h}}{32}$ We can find $D$ when $h=0$: $D = \frac{v^2~sin~\theta~cos~\theta+v~cos~\theta~\sqrt{(v~sin~\theta)^2+(64)(0)}}{32}$ $D = \frac{v^2~sin~\theta~cos~\theta+v~cos~\theta~\sqrt{(v~sin~\theta)^2}}{32}$ $D = \frac{v^2~sin~\theta~cos~\theta+v^2~sin~\theta~cos~\theta}{32}$ $D = \frac{2~v^2~sin~\theta~cos~\theta}{32}$ $D = \frac{v^2~sin~(2~\theta)}{32}$
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