## Engineering Mechanics: Statics & Dynamics (14th Edition)

$\theta=66.9^{\circ}$
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: $\theta=66.9^{\circ}$