Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2016 / Paper 1 / Section B / B8 / d / Solution

At first contact the globally highest trajectory just touches the ceiling at its apex. Its flight time to that point is $t_{\ast }=r\omega \cos {\theta }_{\ast }/g$. The horizontal displacement from the wheel axis isHence the first point struck is directly above the rotation axis, independently of $h$. The release starts to the left of the axis, but the drop's horizontal travel exactly cancels that offset at the optimum apex. The first-contact angular velocity satisfies $r^2{\omega }_{\ast }^2=g(h+\sqrt{h^2-r^2})$.