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

$4.23m$
According to the principle of impulse and momentum in the y-direction, we have: $mv_{y_1}+\Sigma \int ^{t_2}_{t_1} F_y dt=mv_{y_2}$ We plug in the known values to obtain: $10+2\int ^{2}_{0} (100+5t^2)dt-20(9.81)t=20v$ This simplifies to: $v=0.1667t^3+0.19t$ Now the we can determine how high the crate has moved as $v=\frac{ds}{dt}$ $\implies \int ^s _{0} dS=\int ^{t}_{0} (0.1667 t^3+0.19t) dt$ $\implies S=0.0417t^4+0.095t^2$ when $t=3s$, then $S=0.0417(3)^4+0.095(3)^2=4.23m$