Answer
$h_B=0.9797ft$
Work Step by Step
The required height can be determined as follows:
According to the conservation of energy
$T_1+V_1=T_2+V_2$ [eq(1)]
As $T_1=\frac{1}{2}I\omega_1^2=\frac{1}{2}I(0)^2=0J$
$V_1=mgh_1=15(1)=15J$
and $T_2=\frac{1}{2}I\omega_2^2$
$\implies T_2=\frac{1}{2}(\frac{mL^2}{3})\omega_2^2$
$\implies T_2=\frac{mL^2}{6}\omega_2^2=\frac{15(2)^2}{32.2\times 6}\omega_2^2=0.3105\omega_2^2$
$V_2=mgh_2=mg(0)=0J$
We plug in the known values in eq(1) to obtain:
$0+15=0.3105\omega_2^2+0$
This simplifies to:
$\omega_2=6.949 rad/s$
We know that
$e=\frac{(v_g)_3-(v_B)_3}{(v_B)_2-(v_g)_2}$
$\implies 0.7=\frac{0-\omega_3 r}{\omega_2 r-0}$
$\implies \omega_3=4.8643 rad/s$
Now, we apply the conservation of energy for the rebound motion
$T_3+V_3=T_4+V_4$ [eq(2)]
As $T_3=\frac{1}{2}I\omega_3^2$
$\implies T_3=\frac{1}{2}(\frac{mL^2}{3})\omega_3^2$
$\implies T_3=\frac{1}{2}(\frac{15}{32.2})(\frac{(2)^2}{3})(4.8643)^2$
$\implies T_3=7.348J$
$V_3=mgh_3=mg(0)=0J$
$T_4=\frac{1}{2}I\omega_4^2=\frac{1}{2}I(0)^2=0J$
and $V_4=mgh_4=15h_4$
Now we plug in the known values in eq(2) to obtain:
$7.348+0=0+15h_4$
$\implies h_4=0.489ft$
We know that
$\frac{h_4}{L/2}=\frac{hB}{L}$
$\implies h_B=2h_4$
$\implies h_B=2(0.489ft)=0.9797ft$