Answer
The arrow was shot at 29.2 ft/s at an angle of elevation of $59.0^{\circ}$.
Work Step by Step
$|\textbf{V}|=\sqrt {|\textbf{V}_{x}|^{2}+|\textbf{V}_{y}|^{2}}=\sqrt {(15.0\,ft/s)^{2}+(25.0\,ft/s)^{2}}$
$=29.2\,ft/s$
Let the angle of elevation be $\theta$. Then
$\tan\theta=\frac{|\textbf{V}_{y}|}{|\textbf{V}_{x}|}=\frac{25.0}{15.0}$
$\implies \theta=\tan^{-1}(\frac{25.0}{15.0})=59.0^{\circ}$