Answer
$$ v_G=11.9 \mathrm{ft} / \mathrm{s}
$$
Work Step by Step
$$
\begin{aligned}
& v_G=0.5 \omega_G \\
& T_1+\Sigma U_{1-2}=T_2 \\
& 0+10(10.5)\left(1-\cos 45^{\circ}\right)=\frac{1}{2}\left(\frac{10}{32.2}\right) v_G^2+\frac{1}{2}\left[\frac{2}{5}\left(\frac{10}{32.2}\right)(0.5)^2\right]\left(\frac{v_G}{0.5}\right)^2 \\
& v_G=11.9 \mathrm{ft} / \mathrm{s}
\end{aligned}
$$
