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

$v=60m/s$
We can determine the required speed as follows: We apply the principle of impulse and momentum in the $y$-direction: $mv_{y_1}+\Sigma \int_{t_1}^{t_2}F_ydt=mv_{y_2}$ We plug in the known values to obtain: $2(-90)+\frac{1}{2}(0.4\times 10^{-3})(500\times 10^3)=2v$ This simplifies to: $v=60m/s$