Engineering Mechanics: Statics & Dynamics (14th Edition)

$v_B=31.5ft/s$, $d=22.6ft$, $v_C=54.1ft/s$
We can determine the required speed and distance as follows: $T_A+\Sigma U_{A-B}=T_B$ $\implies \frac{1}{2}mv_A^2+Wh_{AB}=\frac{1}{2}mv_B^2$ We plug in the known values to obtain: $\frac{1}{2}(\frac{2}{32.2})(5)^2+(2)(15)=\frac{1}{2}(\frac{2}{32.2})v_B^2$ This simplifies to: $v_B=31.5ft/s$ According to the equation of kinematics $y=y_{\circ}+v_{y_{\circ}}t-\frac{1}{2}gt^2$ We plug in the known values to obtain: $30=0+(31.48)(\frac{3}{5})t+\frac{1}{2}(32.2)t^2$ This simplifies to: $t=0.899s$ We know that $d=x_{\circ}+v_{x_\circ}t=0$ $\implies 0+ (31.48)(\frac{4}{5})(0.899)$ This simplifies to: $d=22.6ft$ Now, according to the work and energy principle $\frac{1}{2}mv_A^2+Wh_{AC}=\frac{1}{2}mv_C^2$ We plug in the known values to obtain: $\frac{2}{32.2}(5)^2+2(45)=\frac{1}{2}v_C^2$ This simplifies to: $v_C=54.1ft/s$