## Engineering Mechanics: Statics & Dynamics (14th Edition)

$28.8ft/s$
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: $v=28.8ft/s$