How Fast is the Skier Traveling Before Landing?

[benji]

Alright, so here's the problem:

An extreme skier, starting from rest, coasts down a mountain that makes an angle of 25.0-degrees with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.200. She coasts for a distance of 10.4 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3.50 m below the edge. How fast is she going just before she lands?

So I didn't want to explain everything I tried, so I just scanned my paper ;). Here it is:

http://img129.exs.cx/img129/6093/physics_problem.gif

Can anyone see what I'm doing wrong? The correct answer is supposed to be 10.9 m/s...

 

[Phymath]

Don't use forces, use energy

the velocity whe the skier leaves the cliff is...
$$\frac{1}{2}mv^2 = mgh$$ 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
$$v_x = v cos 25 \ v_y = v sin 25$$
thus solve for the time to fall the height 3.5 m
$$0m = 3.5 m - v sin 25 t - 1/2 g t^2$$
t= 0.53s

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

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

 

[Leong]

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

 

[benji]

Thanks guys, really appreciate your help.