Skateboarding Down a Ramp: Finding Speed at Bottom

[sallychan]

Homework Statement

A student is skateboarding down a ramp that is 5.9 m long and inclined at 12° with respect to the horizontal. The initial speed of the skateboarder at the top of the ramp is 2.3 m/s. Neglect friction and find the speed at the bottom of the ramp.

Homework Equations

This is my steps:
As I don't know the weight of the student, I don't know the force. So I can't use F-ma

Vo=2.3
Vox= 2.3cos78°= 0.48
Voy= 2.3sin78°=2.25
t=0.32


ΔY (found by SOPHATOA) =Voy.t+1/2 a t^2
1.23= 2.25t+1/2(9.8) t^2

ΔX=Vox.t+1/2 a t^2
5.8=0.48(0.32)+1/2.a.(0.32)
ax=110.3

Vx=Vox+at
Vx=0.48+110.3(0.32)
Vx=35.8

Vy=Voy+at
Vy=2.25+9.8(0.32)
Vy=5.386

therefore the resultant of Vx and Vy is 36.2

The Attempt at a Solution

my answer is wrong

I really need some help :D

 

[Enigman]

Mg is counter-acted by normal force...
For easier methods:
You could just use conservation of energy.
Or you could rotate the coordinate axes.(x-axis along the ramp, y perpendicular)