Answer
$7.208m/s$
Work Step by Step
It is given that $v_1=18Km/h=5m/s$
We know that $S=S_{\circ}+v_{\circ}t -\frac{1}{2}gt^2$
$\implies -10=0+5t-\frac{1}{2}(9.81)t^2$
$\implies t=2.026s$
Now according to the impulse momentum principle
$mv_{y1}+\Sigma \int^{t_2}_{t_1} F_ydt=mv_{y_2}$
We plug in the known values to obtain:
$(400-4)(5)+2.026(2.026)-(400-40)(9.81)(2.026)=(400-40)v_2$
$\implies v_2=7.208m/s$
