## Trigonometry 7th Edition

Magnitude of the velocity $|\textbf{V}|=38.1\,ft/s$ Angle of elevation $\theta=23.2^{\circ}$
$|\textbf{V}|=\sqrt {|\textbf{V}_{x}|^{2}+|\textbf{V}_{y}|^{2}}=\sqrt {(35.0\,ft/s)^{2}+(15.0\,ft/s)^{2}}$ $=38.1\,ft/s$ Let the angle of elevation be $\theta$. Then, $\tan\theta=\frac{|\textbf{V}_{y}|}{|\textbf{V}_{x}|}=\frac{15.0}{35.0}$ $\theta=\tan^{-1}(\frac{15.0}{35.0})=23.2^{\circ}$