Answer
$\theta=66.9^{\circ}$
Work Step by Step
We can determine the required angle as follows:
We know that
$e=\frac{v_{r_2}-v_{d_2}}{v_{d_1}-v_{r_1}}$
$\implies 0.4=\frac{v_{r_1}-0}{0-v_{r_{\circ}}}$
$\implies v_{r_2}=-0.4v_{r_1}$ [eq(1)]
According to the conservation of energy
$\frac{1}{2}m_rv_{\circ}^2+m_rg_{r_{\circ}}=\frac{1}{2}m_rv_{r_1}^2+m_rgh_{r_1}$
We plug in the known values to obtain:
$\frac{1}{2}(\frac{300}{32.2})(0)^2+300(-10sin30)=\frac{1}{2}(\frac{300}{32.2})v_{r_1}^2+300(-10)$
This simplifies to:
$v_{r_1}=17.944m/s$
We plug in this value in eq(1) to obtian:
$v_{r_2}=-0.4v_{r_1}$
$\implies v_{r_2}=-0.4(-17.944)=7.177ft/s$
Now, we apply the conservation of energy after the impact
$\frac{1}{2}m_rv_{r_2}^2+m_rg_{r_2}=\frac{1}{2}m_rv_{r_3}^2+m_rgh_{r_3}$
We plug in the known values to obtain:
$\frac{1}{2}(\frac{300}{32.2})(7.177)^2+300(-10)=\frac{1}{2}(\frac{300}{32.2})(0)^2+300(-10sin\theta)$
This simplifies to:
$\theta=66.9^{\circ}$
