Answer
$h$ would be the same
Work Step by Step
Lets calculate total energy of skier relative to end of ramp (initial kinetic energy of skier is zero since it started with zero speed) $E_{tot}=E_{k}+E_{p}=0+mgH=mgH(1)$ At the end of the ramp the potential energy of skier will be zero ( becouse it will be at the same height with end of ramp) so it will only have kinetik energy $E_{tot}=E_{k}+E_{p}=E_{k_{1}}+0=E_{k_{1}}\left( 2\right) $ and also we can write: $E_{k_{1}}=\dfrac {mv^{2}_{1}}{2}\left( 3\right) $ $v_{1}$ is the speed at the end of the ramp. So from (1),(2) and (3) we get $\dfrac {mv^{2}_{1}}{2}=mgH\Rightarrow v_{1}=\sqrt {2gH}(4)$ So lets calculate maximum height can skier go: $h=\dfrac {v^{2}_{0}\sin ^{2}\theta }{2g}\left( 5\right) $ So from (4) and (5) we get: $h=H\sin ^{2}\theta (6)$ As you see from equation (6) maximum height skier can go $h$ Doesnt depend on mass of skier ( there is no $m$ in this equation$ so $h$ would be the same
