Answer
$v=0.301m/s$
Work Step by Step
We can find the required velocity as
$2F-W=0$
$\implies 2(625t^2)=mg=(400)(9.81)$
$\implies t=1.772s$
We know that
$\Sigma F_y=0$
$\implies 2F-W=0$
$\implies 2(625t^2)-(400)(9.81)=400a$
$\implies a=3.125t^2-9.81$
Now, $v=\int_{1.772}^t(3.125t^2-9.81)dt$
$\implies v=(\frac{3.125 t^3}{3}-9.81t)|_{1.772}^t$
This simplifies to:
$v=0.301m/s$