Answer
The skier lands 1.11 meters below the launch level.
Work Step by Step
We can use the equation $y = tan\theta * x - gx^{2}/2(vcos\theta)^{2}$
Let d be the distance of the slope, so that $y = d sin a$ and $x = d cos a.$
Substitute those values and let $a = 9.0^{\circ}, \theta = 11.3^{\circ}$, and $v = 10 m/s$.
Solve for d to get that d = 7.117m.
Plug that back into $y = d sin a = 7.117 sin 9.0^{\circ}$ and you get that the distance change in y is -1.11m, or that the skier ends up 1.11 meters below the launch level.
