#### Answer

$10.1ft/s$

#### Work Step by Step

The required speed can be determined as follows:
$F=30+t^2=34$
$\implies t=2s$
Now we apply impulse momentum principle
$mv_1+\Sigma \int ^{t_2}_{t_1} F_y dt=mv_2$
$\implies \frac{34}{32.2}(0)+\int ^{4}_{2} (30+t^2)dt-W\Delta t=(\frac{34}{32.2})v$
This simplifies to:
$v=10.1ft/s$