#### Answer

$h=18m$

#### Work Step by Step

1) Find the speed of the skier so that the skier just loses contact with the snow at the crest of the 2nd hill.
The skier starts to lose contact with the snow at that point when there is no longer normal force $F_N$, leaving gravitational force to provide the only centripetal force there: $$mg=F_c=\frac{mv^2}{r}$$ $$g=\frac{v^2}{r}$$ $$v=\sqrt {gr}=18.78m/s$$
2) Now we apply the principle of energy conservation over the whole road
We have $$\Delta KE = -\Delta PE$$ $$\frac{1}{2}m(v_f^2-v_0^2)=-mg\Delta h$$ $$(v_f^2-v_0^2)=-2g\Delta h$$
We know $v_0=0$. We take the finish point to be at the crest of the 2nd hill, so $v_f=18.78m/s$ and $\Delta h=h_f-h_0=-h$
$$18.78^2=-2\times9.8\times (-h)$$ $$h=18m$$