Super fun projectile motion ski slope problem

[Peppo]

Homework Statement

A ski-jumper comes off the bottom of the slope going horizontally at a speed of 20 ms. It starts off 3 m above a slope going downwards at an angle of 45 degrees. How far along the slope does it land.

Homework Equations

1) y= Vsinθ - (.5)gt^2
2) x= Vcosθt
3) tanβ = ||y|| / ||x||

The Attempt at a Solution

I attempted to find the solution by considering the beginning of the slope to be at the origin with the skiier starting 3 metres above it so my equation for y position was:

4) y = Vsinθt - (.5)gt^2 + 3

-which simplifies to: y = -.5gt^2 + 3 since sinθ is zero.

The x position is given by equation 2.
Using equation 3 to change equation 2 to give the y-position corresponding to the x-position led me to:

-y/V = t (since Vcosθ = V in this case)

I then used this in equation 4 to replace t, and used the quadratic formula to get an answer for the y value.

After this step the distance is just a matter of finding the hypotenuse of a triangle and isn't of interest to me, but what I'm concerned about is that the answer I got is off slightly from what I found when I measured out the problem on computation paper and it's outside (just by a little too much) the error margins given the tools I used for drawing it.

My question is whether or not I correctly incorporated the starting position of 3 metres above the 45 degree angle slope? Thanks :)

 

[NascentOxygen]

That looks right. If you'd shown your working and answers I would have seen whether your answers accord with mine.