let \(\displaystyle \L \theta\) = angle formed at the hemispere's center between the top of the hemisphere and the boy's position as he slides down the ice.
two forces acting on the boy ... normal force perpendicular to the ice and the boy's weight.
the boy's weight has a radial component = \(\displaystyle \L mg\cos{\theta}\), and
a tangential component = \(\displaystyle \L mg\sin{\theta}\).
as the boy slides down and picks up speed, the radial component of weight is greater than the normal force of the ice, providing the necessary centripetal force to keep the boy moving in a circular path. at the point where the normal force equals the radial component of weight, the centripetal force equals zero ... the result is the boy does not have the necessary centripetal force to remain moving in a circular path ... he "flies" off the hemisphere's surface.
\(\displaystyle \L mg\cos{\theta} = \frac{mv^2}{R}\)
solving for \(\displaystyle \L v^2\) ...
\(\displaystyle \L Rg\cos{\theta} = v^2\)
since no energy is "lost" to friction, the change in the boy's potential energy is all converted to kinetic energy ...
\(\displaystyle \L mg(R - h) = \frac{mv^2}{2}\)
\(\displaystyle \L 2g(R - h) = v^2\)
setting the \(\displaystyle v^2\)'s equal ...
\(\displaystyle \L Rg\cos{\theta} = 2g(R - h)\)
since \(\displaystyle \L h = R\cos{\theta}\) ...
\(\displaystyle \L gh = 2g(R - h)\)
\(\displaystyle \L h = 2(R - h)\)
\(\displaystyle \L 3h = 2R\)
\(\displaystyle \L h = \frac{2R}{3}\)
