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Uniform Acceleration/Projectile Motion

  1. Oct 2, 2007 #1
    1. The problem statement, all variables and given/known data
    "A long jumper's jump measures 7.5m, and he is in the air for 0.8s. His centre of gravity falls 0.95m between his take off and landing."

    a) Calculate his horizontal velocity at take-off.
    b) Show that his vertical velocity is 2.7(3) m/s.
    c) Hence calculate the angle at which he projects himself at take-off.

    2. Relevant equations

    [tex] v_{ave} = \Delta x / \Delta t [/tex]
    [tex] v_{horiz.} = v_{result.}cos \Theta [/tex]
    [tex] v_{vert.} = v_{result.}sin \Theta [/tex]

    There may be some more that have to come into use, too...

    3. The attempt at a solution

    I calculated that his horizontal velocity at take-off is 9.38m/s, since he travels 7.5m horizontally in 0.8s.

    I have spent about an hour trying to work out what to calculate, and how, in order to be able to calculate the vertical velocity at take-off.

    I tried to calculate his instantaneous velocity at the point where his centre of gravity had reached its original point, but I couldn't since I don't know the angle at which he was projected...

    It all goes downhill from here, really.

    Any help would be greatly appreciated.

    Thanks a lot.
  2. jcsd
  3. Oct 2, 2007 #2


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    d = v1*t + (1/2)at^2

    apply this equation for the vertical direction. you have t, a and d. you can solve for v1.
  4. Oct 2, 2007 #3
    Oh, of course... the centre of gravity thing threw me a bit. Thanks!
    Last edited: Oct 2, 2007
  5. Oct 2, 2007 #4


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    Homework Helper

    No prob. just use d = -0.95m
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