Trouble with Proof of Optimal Launch Angle

[stoffer]

Here is my problem, can anyone help me out?

Prove the maximum range of a skier going down a ski jump is given by
(theta) = 45degrees - phi/2 , where theta = optimal launch angle , phi = slope angle of take off with respect to the incline of the hill.

I have the following equations to use

Xf = Vi*cos(theta)*t = d*cos(phi)
Yf = Vi*sin(theta)*t - (1/2)g*t^2 = -d*sin(phi)

Where, Xf= final x component, Yf= final y component, Vi = velocity at launch, t= time jumper is in air, g= grav. constant, d = distance travelled along the inlcine of the hill

I am trying to eliminate t in the above equations and then differentiate to maximize d in terms of theta.

Any ideas how to do this? What is a good first step? Any help you can give will be greatly appreciated.
Thanx

[stoffer]

phi = angle the hill makes with the horizontal

[NateTG]

WARNING: This post contains technical errors, but it should still be instructive.

If you treat this as a system of a parabola and a line, you get the equations:
The slope:
y=mx
and
the skiers' path:
y=-(x-k)^2+l=-x^2+2kx-k^2+l
Substitution yields
y=-m^2y^2+2kmy-k^2+l
so
y=\frac{2km+\sqrt{4k^2m^2+4m^2(l-k^2)}}{2m^2}.
Which can obviously be simplified to:
y=\frac{k+\sqrt{l}}{m}

Now, you need to find m,k, \mbox{and } l.

m should be pretty straightforward to get.

To find k and l you need to realize that the skier reaches his peak when x-k=0 so k and l are the horizontal and vertical distances to the *top* of the skiers jump - which you should be able to find easily.