A particle P is projected horizontally with speed u, where u^2 < 2ag, from the highest point A of a fixed smooth sphere of radius a and whose centre is at O. The particle slides on the outer surface of the sphere. Show that it leaves the sphere when PO makes an angle arccos ($\displaystyle \frac{2}{3}+\frac{u^2}{3ag}$) with the vertical.

I figured out that when speed=2ag, reaction will be 0 and the particle will leave the sphere. And the change in speed have something to do with the energy change mgh=.5 mv^2. But all my workings to find out h lead to very complicated expressions. So how do i prove the angle is arccos ($\displaystyle \frac{2}{3}+\frac{u^2}{3ag}$) ?

Please help me associate the change in speed with the angle.

(vertical circular motion)