#### Answer

$v_B=31.5ft/s$, $d=22.6ft$, $v_C=54.1ft/s$

#### Work Step by Step

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:
$v_C=54.1ft/s$