# Trouble with Proof of Optimal Launch Angle

1. Jan 22, 2004

### 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

2. Jan 22, 2004

### stoffer

minor correction

phi = angle the hill makes with the horizontal

3. Jan 22, 2004

### 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.

Last edited: Jan 22, 2004