Max value of v

Let v_i be the initial velocity and v_f be the final velocity.

When it is at the edge (where the plane bends down) the ball rotates about that point and then continues to move down the incline.

While rotating,

\frac{mv_f^2}{r}=mg\sin\alpha -N

From energy conservation we get,

\frac{1}{2}I\frac{v_f^2}{r^2}-\frac{1}{2}I\frac{v_i^2}{r^2}=mgr(1-\sin \alpha)

From parallel axis theorem we get I = 12mr²+mr²=32mr²

When the ball is just about to jump N=0.

N=0\rightarrow \frac{v_f^2}{r}=g\sin\alpha

\frac{3m}{4}(v_f^2-v_i^2)=mgr(1-\sin \alpha)

\rightarrow v_i^2=gr\sin\alpha -\frac{4}{3}gr(1-\sin\alpha)
\rightarrow v_i=v_{max}=\sqrt{\frac{gr}{3}(7\sin\alpha - 4)}