The normal contact impulse $J$ between the rod and plane has two effects : it changes the velocity of the CM of the rod, and it makes the rod rotate about its CM.

The impulsive force is related to the change in momentum of the CM of the rod, and the impulsive torque to the change in angular momentum of the rod, as follows : $$J=m(v_0+v_1)$$ $$J\frac{L}{2}\cos\theta=I\omega$$ Here $v_0, v_1$ are the vertical velocities of the CM before and after the collision, and $I=\frac{1}{12}mL^2$ is the moment of inertia of the rod about its CM.

The rod is not rotating initially. After the collision, the velocity of the end A which collided with the ground is $v_2'=\frac{L}{2}\omega$ relative to the CM of the rod, directed perpendicular to the rod, ie at an angle of $\theta=60^{\circ}$ to the vertical. The CM of the rod is moving upwards with speed $v_1$ so the vertical component of the velocity of A relative to the ground after collision is $$v_2=v_2'\cos\theta+v_1$$

The elasticity of the collision is known ($e=1$) so the **Law of Restitution** can be applied to the relative velocities of approach and separation at the point of contact : $$v_2=ev_0$$

From the above equations you can eliminate $v_2, v_2', \omega$ to find the rebound speed $v_1$ of the CM of the rod and thereby the height to which the CM rises after collision.