Answer
$v_c=17.7ft/s$
Work Step by Step
We can determine the required speed as follows:
According to the conservation of energy equaiton
$\frac{1}{2}m_Bv_{B1}^2+\frac{1}{2}m_Cv_{C_1}^2+V_{B_1}+V_{C_1}=\frac{1}{2}m_Bv_{B_2}^2+\frac{1}{2}m_Cv_{C_2}^2+V_{B_2}+V_{C_2}$
We plug in the known values to obtian:
$0+0+0+0=\frac{1}{2}(\frac{200}{32.2})(\frac{-v_C}{2})^2+\frac{1}{2}(\frac{600}{32.2})(v_C)^2+200(15)+600(-30sin20)$
This simplifies to:
$v_c=17.7ft/s$
