Answer
$v=3.62m/s\uparrow$
Work Step by Step
We can determine the required velocity as follows:
$\Sigma F_y=0$
$\implies F+F-T=0$
$\implies T=2F$
$\implies T=2(150)=300N$
For block A, $\Sigma F_y=ma$
$\implies 2F-W=ma$
$\implies a=\frac{2T}{m}-g$
We plug in the known values to obtain:
$a=\frac{2(300)}{50}-9.81$
$\implies a=2.19$
Now, $v^2=v^2_{\circ}+2a(y-0)$
We plug in the known values to obtain:
$v=\sqrt{0+0(2.19)(3)}$
This simplifies to:
$v=3.62m/s$