#### Answer

$d=36.2ft$

#### Work Step by Step

According to the principle of work and energy
$\frac{1}{2}mv_A^2-WS+\frac{1}{2}kS^2=\frac{1}{2}mv_B^2$
We plug in the known values to obtain:
$0-10(3)+\frac{1}{2}(100)(2)^2=\frac{1}{2}(\frac{10}{32.2})v_B^2$
This simplifies to:
$v_B=33.09ft/s$
From the equations of kinematics
$S_y=S_{y_{\circ}}+v_{y_{\circ}}t+\frac{1}{2}a_ct^2$
We plug in the known values to obtain:
$-3=0+(33.09)(\frac{3}{5})t-\frac{1}{2}(32.2)t^2$
$\implies t=1.369s$
Now $d=S_x=S_{x_{\circ}}+v_{x_{\circ}}t$
We plug in the known values to obtain:
$d=0+33.09(\frac{4}{5})(1.369)$
This simplifies to:
$d=36.2ft$