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Dynamics/Kinematics - Jumper beats record jump

  1. Nov 4, 2013 #1
    1. The problem statement, all variables and given/known data
    The jumper is 60 kg and wants to beat the school's record jump, 1.1 m. She bends her knees a distance of .5 m before jumping. How many Newtons must Fjump be to for her to beat the record?

    2. Relevant equations
    ma = ƩF
    V2 - V02 = 2a(X - X0)

    My teacher stated I would just need these two equations, and that I would use the latter twice.
    3. The attempt at a solution
    Jumping up to the apex:
    02 - 02 = (2)(-10)(1.1)
    She jumps up from rest and the velocity of a mass at the apex is zero; right after she jumps, the only force acting on her is gravity, which is -10 m/s2. I assume 1.1 to be X because she can beat the previous runner's score if she jumps 1.1000001 m, essentially 1.1. So in my solution I could say "she has to jump with a force great than _X."

    60a = Fjump - 600
    I need acceleration from the previous step.

    a) I don't know how to work in the .5 m into the equations. Would I add .5 to X? I would still get a nonsensical answer for step 1.
    b) What is wrong with step 1's set up?
     
  2. jcsd
  3. Nov 4, 2013 #2

    cepheid

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    The problem with step 1 is with your definition of "initial". The v0 in your equation is not the speed at the beginning of the jump. It's the launch speed: the speed at the end of the jump, once the jumper's legs are straight and she leaves the ground. And v is the final speed, the speed at the apex, which is 0. So, we have:

    v0^2 = -2gh, where h = 1.1 m, the desired height of the jump.

    Okay, that's all well and good, and from this we can find the launch speed needed.

    How do we get the jump force? That's step 2, where we apply the kinematics equation again. It must be that the acceleration from the net force Fjump - weight, applied over a distance of 0.5 m, gives you a final v that is equal to v0 from step 1. So, in *step 2*, "initial" now means at the beginning of the jump. So v0 = 0, and v = the needed launch speed from step 1, and d = 0.5 m. From this you can solve for 'a', which gives you Fnet.
     
  4. Nov 5, 2013 #3
    Cepheid variable!
    ---
    Straight legs:
    v0^2 = 2*-10*1.1
    v0 = 4.69

    Bent:
    4.69^2 - 0^2 = 2*.5*a
    a = 21.996

    60*22 = F(jump) - 600
    F(jump) = 1920 N
    ---
    Thanks!
     
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