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