Answer
$h = \frac{2R}{3}$
Work Step by Step
We can use conservation of energy to find an expression for $v^2$, where $v$ is the speed when the skier leaves the surface:
$\frac{1}{2}mv^2+mgh = mgR$
$\frac{1}{2}mv^2 = mg~(R-h)$
$v^2 = 2g~(R-h)$
Let $\theta$ be the angle above the horizontal, from the center of the half dome, when the skier leaves the surface. Note that $sin~\theta = \frac{h}{R}$. At this moment, the component of the skier's weight directed into the surface of the slope is equal in magnitude to the required centripetal force:
$mg~sin~\theta = \frac{mv^2}{R}$
$v^2 = gR~sin~\theta$
$v^2 = (gR)~(\frac{h}{R})$
$v^2 = gh$
We can equate the two expressions for $v^2$:
$gh = 2g~(R-h)$
$h = 2~(R-h)$
$3h = 2R$
$h = \frac{2R}{3}$
