Solution

Angular momentum

$$\left\{\begin{matrix} L=m_2v\frac{l}{2} \\ L=I\omega \end{matrix}\right.$$

Parallel axis theorem

$$I=I_0+m_2d^2=\frac{1}{3}m_1l^2+m_2\left(\frac{l}{2}\right)^2$$

$$m_2v\frac{l}{2}=\omega\left(\frac{1}{3}m_1l^2+\frac{1}{4}m_2l^2\right)$$

From where

$$\omega = \frac{6m_2}{4m_1+3m_2}\frac{v}{l}$$

From the figure

$$h=\frac{l}{2}(1-\cos\alpha )$$

Law of conservation of energy in rotational motion

$$\frac{I\omega^2}{2}=(m_1+m_2)\frac{l}{2}(1-\cos\alpha )$$

Expressing $\cos\alpha$

$$1-\cos\alpha =\frac{I\omega^2}{(m_1+m_2)l}$$

$$\boxed{\cos\alpha =1-\frac{3m_2^2v^2}{gl(4m_1+3m_2)(m_1+m_2)}}$$