Solution

Stage 1. Motion with the Block in Contact

A. Velocity relationship

Because the wedge slides against a frictionless vertical wall while its slanted face remains in contact with the block, we can write the relation between the speed of the wedge and the speed of the block:

\begin{equation*} V_b = V_w \cot\alpha \end{equation*}
This relation holds throughout the period when the wedge and block remain in contact.

B. Conservation of energy

\begin{equation*} Mgh = \frac{1}{2}MV_w^2 + \frac{1}{2}m(V_w \cot\alpha)^2 \end{equation*}

Solve for $V_w$:

\begin{equation*} V_w^2 = \frac{2Mgh}{M + m\cot^2\alpha} \end{equation*}

Stage 2. Free Fall After Separation

After separation, the wedge is no longer in contact with the block and is free to fall from its current height. In order to reach the floor, the wedge must fall an additional distance of $(H - h)$. Using the kinematic relation for free fall:

\begin{equation*} v_M^2 = V_w^2 + 2g(H - h) \end{equation*}

After substituting the expression for $V_w$ we get:

\begin{equation*} v_M = \sqrt{\frac{2Mgh}{M + m\cot^2\alpha} + 2g(H - h)} \end{equation*}