the velocity whe the skier leaves the cliff is...
[tex]\frac{1}{2}mv^2 = mgh[/tex] while h = 10.4m sin 25
v = sqrt(2g(10.4 sin 25)) = 9.29 m/s

the skiers velocity is 25 degrees below the horizontal, thus
[tex]v_x = v cos 25 \ v_y = v sin 25 [/tex]
thus solve for the time to fall the height 3.5 m
[tex] 0m = 3.5 m - v sin 25 t - 1/2 g t^2[/tex]
t= 0.53s

and when the skier makes impact the velocity is
[tex]v_y = v sin 25 + 9.8 (0.53 s)[/tex]
[tex]v_x = v cos 25[/tex]
so the total velocity at impact is..
[tex] v = sqrt( (v_x)^2+(v_y)^2) = 12.4128 m/s[/tex]

so i believe your answers wrong doing this is also very complicated but i thought would make u believe the answer more cause I use projectiles, gravity is a conservative force so therefore it doesn't matter what path the skier takes to drop the height u can find the velocity by 1/2mv^2 = mgh
where h is (10.4m sin 25 + 3.5m) the total height from the ground. you'll get the same result.

edit: i forgot the coff of friction, however use work engery principle in the same way i did this problem that might help

Your sum of all x component forces :
[tex]F_{g}sin25-F_{fricition}=ma_x[/tex]
Your sum of all y component forces :
[tex]\Sigma F_{y}=ma_y[/tex]
N-mgcos25=0