Answer
$r_2=1.52ft$, $t=0.74s$
Work Step by Step
The required time and distance can be determined as follows:
We know that
$v_2=\sqrt{(v_{\theta})_2^2+(v_r)_2^2}$
$12=\sqrt{(v_{\theta})^2+(2)^2}$
$\implies v_{\theta}=11.83ft/s$
According to the conservation of angular momentum
$m_B(v_B)_1r_1=m_B(v_{\theta})r_2$
$\implies 6\times 3=11.83r_2$
$\implies r_2=1.52ft$
Now, $t=\frac{\Delta r}{v_r}$
We plug in the known values to obtain:
$t=\frac{3-1.52}{2}$
This simplifies to:
$t=0.74s$
