#### Answer

$4.047m/s$

#### Work Step by Step

The required speed can be calculated as
$3F=W$
$\implies 3(10t^2+300)=100(9.81)$
$\implies t=1.643s$
Now,
$mv_{y1}+\Sigma \int _{t_1}^{t_2}F_ydt=mv_{y2}$
We plug in the known values to obtain:
$0+\int_{1.643}^4 3(10t^2+300)dt-100(9.81)(4-1.643)=100v$
$\implies v=4.047m/s$