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Homework Help: Trouble with Proof of Optimal Launch Angle

  1. Jan 22, 2004 #1
    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. jcsd
  3. Jan 22, 2004 #2
    minor correction

    phi = angle the hill makes with the horizontal
     
  4. Jan 22, 2004 #3

    NateTG

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    Science Advisor
    Homework Helper

    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:
    [tex]y=mx[/tex]
    and
    the skiers' path:
    [tex]y=-(x-k)^2+l=-x^2+2kx-k^2+l[/tex]
    Substitution yields
    [tex]y=-m^2y^2+2kmy-k^2+l[/tex]
    so
    [tex]y=\frac{2km+\sqrt{4k^2m^2+4m^2(l-k^2)}}{2m^2}[/tex].
    Which can obviously be simplified to:
    [tex]y=\frac{k+\sqrt{l}}{m}[/tex]

    Now, you need to find [tex]m,k, \mbox{and } l[/tex].

    [tex]m[/tex] should be pretty straightforward to get.

    To find [tex]k[/tex] and [tex]l[/tex] you need to realize that the skier reaches his peak when [tex]x-k=0[/tex] so [tex]k[/tex] and [tex]l[/tex] 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
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