Answer
$$
v_A=4.00 \mathrm{~m} / \mathrm{s}
$$
Work Step by Step
$$
\begin{aligned}
& x^2+y^2=5^2 \\
& x^2+(7-y)^2=4^2 \\
& \text { Thus, } \quad y=4.1429 \mathrm{~m} \\
& \qquad \quad x=2.7994 \mathrm{~m} \\
& (5)^2=(4)^2+(7)^2-2(4)(7) \cos \phi \\
& \phi=44.42^{\circ} \\
& h^2=(2)^2+(7)^2-2(2)(7) \cos 44.42^{\circ} \\
& h=5.745 \mathrm{~m} \\
& T_1+V_1=T_2+V_2 \\
& 0+147.15(2)=\frac{1}{2}\left[\frac{1}{12}(15)(4)^2\right] \omega^2+\frac{1}{2}(15)(5.745 \omega)^2+147.15\left(\frac{7-4.1429}{2}\right) \\
& \omega=0.5714 \mathrm{rad} / \mathrm{s} \\
& v_A=0.5714(7)=4.00 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
