#### 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