Answer
$28.8ft/s$
Work Step by Step
According to impulse momentum principle
$mv_1+\Sigma \int ^{t_2}_{t_1}F_xdt=mv_2$
We plug in the known values to obtain:
$\frac{200}{32.2}+\int ^5_{0} 20(t+1) cos 30^{\circ} dt=(\frac{200}{32.2})v$
$\implies -124.22+20 cos 30^{\circ} (0.5t^2+t)|^5_{0}=6.211v$
This simplifies to:
$v=28.8ft/s$
