#### Answer

$R=2.83m$, $v_C=7.67m/s$

#### Work Step by Step

According to the principle of work and energy
$\frac{1}{2}mv_A^2+U_w=\frac{1}{2}mv_B^2$
$\implies \frac{1}{2}mv_A^2+W(3-2)=\frac{1}{2}mv_B^2$
We plug in the known values to obtain:
$0-(0.005)(9.81)(1)=\frac{1}{2}(0.005)v_B^2$
$\implies v_B=4.429m/s$
From the equations of kinematics, we get
$S_y=S_{\circ}+v_{y_{\circ}}t+\frac{1}{2}gt^2$
$\implies 2=0+0+\frac{1}{2}(9.81)t^2$
$\implies t=0.639s$
Now $S_x=R=S_{x_{\circ}}+v_{x_{\circ}}t$
$\implies R=(4.429)(0.639)$
$\implies R=2.83m$
We know that
$\frac{1}{2}mv_A^2+U_w=\frac{1}{2}mv_C^2$
We plug in the known values to obtain:
$0+(0.005)(9.81)(3)=\frac{1}{2}(0.005)v_C^2$
This simplifies to:
$v_C=7.67m/s$