The track consists of a concave part (valley) AB and a convex part (hump) BC. The angular span of AB is $120^\circ$. Assuming symmetry for the valley, the bottom-most point is the midpoint of the arc. Thus, point B is located at an angle of $60^\circ$ from the vertical passing through the center of curvature of AB.

For the convex part BC (hump), the motion begins at B. The tangent is continuous, so at B, the track slopes upwards at $60^\circ$. The hump curves downwards from there. The height of point B above the bottom of the valley (Lowest Point $L$) is: $$ H_B = r(1 – \cos 60^\circ) = r(1 – 0.5) = 0.5r $$

The block P will “certainly lose contact” if it loses contact at the most critical point on the convex track BC. The hint suggests that the minimal speed case corresponds to losing contact immediately upon entering the convex section at B.

At any point on the convex track with angle $\phi$ to the vertical, the equation of motion along the normal is: $$ mg \cos \phi – N = \frac{mv^2}{r} $$ To lose contact, $N \le 0$. The limiting condition at point B (where the angle with the vertical for the hump’s center of curvature is also $60^\circ$) is: $$ mg \cos 60^\circ = \frac{mv_B^2}{r} $$ $$ g(0.5) = \frac{v_B^2}{r} \implies v_B^2 = 0.5rg $$

Let $h$ be the release height above the bottom of section AB. Applying conservation of energy from release point to point B: $$ PE_{start} = PE_B + KE_B $$ $$ mgh = mg(H_B) + \frac{1}{2}mv_B^2 $$ Substitute $H_B = 0.5r$ and $v_B^2 = 0.5rg$: $$ mgh = mg(0.5r) + \frac{1}{2}m(0.5rg) $$ $$ h = 0.5r + 0.25r $$ $$ h = 0.75r $$