Question

Determine the height of the surface if the block leaves at point c

Answer 1

A small body of mass mm slides down from the top of a hemisphere of radius rr. There is no friction between the surface of the block and the hemisphere. The height at which the body loses contact with the surface of the sphere is?

This is how I understood the problem: 
First of all, the mass does not lose contact with the surface of the sphere at all, considering it is undergoing centripetal acceleration, where the centripetal force is provided by the component of the mass's weight towards the centre of the hemisphere (mgcosθmgcosθ, where θθ is the angle between the vector of the mass's weight and its component acting towards the centre).
Therefore, since only mgcosθmgcosθ is responsible for the centripetal force, I can form a relationship like this:

mgcosθ = mv2rmgcosθ = mv2r

v = rgcosθ−−−−−√v = rgcosθ

Taking 'h' as the height of the mass from the base of the hemisphere.

cosθ = hrcosθ = hr

Then the velocity of the mass becomes:

v = gh−−√v = gh

The component of the mass's weight along the centre disappears only when θθ becomes 9090 degrees. At this point, it leaves the surface of the hemisphere.
Now, the energy of the mass at the topmost point is:

P.E = mgrP.E = mgr