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The End of the Ski Jump - Optimizing Launch Angle

  1. Jan 17, 2016 #1
    1. The problem statement, all variables and given/known data
    A ski jumper leaves the ski track moving in the horizontal direction with a speed of 25.0 m/s as shown in Figure 4.14. The landing incline below her falls off with a slope of 35.0°. Where does she land on the incline? I've attached an image of the problem, my work is below it.
    RN8sQ5y.png
    yzjFy3T.png


    2. Relevant equations
    xf = (Vicos(theta))t = dcos(phi)
    yf = (Visin(theta)t) - (0.5)gt^2 = -dsin(phi)

    3. The attempt at a solution
    I understand and reached the solution of part a), but I am struggling with part b). It says to eliminate time "t" and differentiate and maximize "d" in terms of angle "theta".

    this is the furthest I've gotten:

    xf = (Vicos(theta))t

    t = xf/(Vicos(theta))

    yf = (Visin(theta))t - (0.5)gt^2

    plugging in t = xf/(Vicos(theta))

    yf = (Visin(theta))(xf/(Vicos(theta)) - (0.5)g(xf/(Vicos(theta))^2
    = xftan(theta) - (0.5)g(xf/(vicos(theta))^2
    = xftan(theta) - xf^2(1/(2gVi^2cos^2(theta))

    (factoring out and dividing by xf)

    yf/xf = tan(theta) - xf/(2gVi^2cos^2(theta))

    -dsin(phi)/(dcos(phi)) = tan(theta) - xf/(2gVi^2cos^2(theta))

    -tan(phi) = tan(theta) - xf/(2gVi^2cos^2(theta))

    I don't know what to do after this step. Any help would be greatly appreciated. Thank you!




     
  2. jcsd
  3. Jan 17, 2016 #2

    haruspex

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    What does theta represent in that equation? Think carefully.
     
  4. Jan 17, 2016 #3
    Hi Haruspex, I believe it represents the launch angle
     
  5. Jan 17, 2016 #4

    haruspex

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    Yes, but angle to what?

    Edit: sorry, forget it..... may have misread something. More soon....
     
  6. Jan 17, 2016 #5
    The angle in respect to the x-axis? Is it because it's initially at zero? Because φ is 35 degrees ... ?
     
  7. Jan 17, 2016 #6
    Oh okay, all good. Thanks for replying to my thread, though!
     
  8. Jan 17, 2016 #7

    haruspex

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    Check that step. The second line is dimensionally inconsistent.
    (This is a useful technique for finding algebraic errors in physics. Check whether the final equation makes sense dimensionally. If it doesn't, do a 'binary chop' to see where it went wrong.)
     
  9. Jan 17, 2016 #8
    Okay thanks, I'll review my work once I'm home. Also, what do you mean by "binary chop"?
     
  10. Jan 17, 2016 #9

    haruspex

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    Say you did eight operations on an equation in sequence. The first equation passes some check (such as dimensional consistency) but the ninth doesn't. so check the one in the middle. If that's ok, check the one half way from there to the end, etc.
    You might find this useful: https://www.physicsforums.com/insights/frequently-made-errors-equation-handling
     
  11. Jan 17, 2016 #10
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