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

$7.208m/s$
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: $(400-4)(5)+2.026(2.026)-(400-40)(9.81)(2.026)=(400-40)v_2$ $\implies v_2=7.208m/s$