Answer
$4.23m$
Work Step by Step
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
$S=0.0417(3)^4+0.095(3)^2=4.23m$
