#### Answer

$6ft/s$

#### Work Step by Step

We apply the principle of impulse and momentum in the x-direction
$mv_{x_1}+\Sigma \int_{t_1}^{t_2} F_x dt=mv_{x_2}$
We plug in the known values to obtain:
$-(\frac{10}{32.2})(2)(3)-T+(0.15)(10)=\frac{10}{32.2}v_{y_2}~~~~$eq(1)
Now, we apply the principle of impulse and momentum in the y-direction
$mv_{y_{1}}+\Sigma \int_{t_1}^{t_2}F_ydt=mv_{y_2}$
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)
Solving eq(1) and eq(2), we obtain:
$T=1.5lb$
and $v_{y_2}=6ft/s$