#### Answer

$10.5ft/s$

#### Work Step by Step

We can determine the required velocity as follows:
First, we apply the principle of impulse and momentum in the x-direction
$mv_{x_1}+\Sigma \int ^{t_2}_{t_1} F_x dt=mv_{x_2}$
We plug in the known values to obtain:
$-(\frac{10}{32.2})(2)(3)-T=\frac{10}{32.2}v_{y_2}$.....eq(1)
Now, applying the principle of impulse and momentum in the y-direction, we get
$mv_{y_1}+\Sigma \int ^{t_2}_{t_1}F_y dt=mv_{y_2}$
We plug in the known values to obtain:
$(\frac{3}{32.2})(3)-2T+3=-\frac{3}{32.2}\frac{v_{y_2}}{2}$....eq(2)
After solving eq(1) and eq(2), we obtain:
$v_{A_2}=10.5ft/s$