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

$10.5ft/s$
We can determine the required velocity as follows: First, we apply the principle of impulse and momentum in the x-direction $mv_{x_1}+\Sigma \int ^{t_2}_{t_1} F_x dt=mv_{x_2}$ We plug in the known values to obtain: $-(\frac{10}{32.2})(2)(3)-T=\frac{10}{32.2}v_{y_2}$.....eq(1) Now, applying the principle of impulse and momentum in the y-direction, we get $mv_{y_1}+\Sigma \int ^{t_2}_{t_1}F_y dt=mv_{y_2}$ We plug in the known values to obtain: $(\frac{3}{32.2})(3)-2T+3=-\frac{3}{32.2}\frac{v_{y_2}}{2}$....eq(2) After solving eq(1) and eq(2), we obtain: $v_{A_2}=10.5ft/s$