#### Answer

$15.6m/s$

#### Work Step by Step

We apply the principle of impulse and momentum as follows:
$mv_{x1}+\Sigma \int_{t_1}^{t_2} F_x dt=mv_{x2}$
We plug in the known values to obtain:
$(2500)(5.556)+\int_{0}^{5} 600t^2 dt=2500v_2$
This simplifies to:
$v_2=15.6m/s$