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A momentum physics question

  1. Nov 20, 2006 #1
    The "force platform" is a tool that is used to analyze the performance of athletes by measuring the vertical force as a function of time that the athlete exerts on the ground in performing various activities. A simplified force vs time graph for an athlete performing a standing high jump is shown below (see attachment). The athlete started the jump at t = 0.0s. How high did this athlete jump?

    My work so far:
    okie, using the equation, I = p:
    Ft = mvf - mvi
    At the highest point, vf = 0, so...
    1000N * 1.0s = -mvi
    Now what do I do? I got an equation with two unknowns. I know that once you solve for vi, you can just use the equation: x = vit + 0.5at^2 to solve for x. The question is how do I get vi?

    If you can't see the attachment, go to: http://students.washington.edu/cy1126/Physics.jpg
     

    Attached Files:

    Last edited: Nov 20, 2006
  2. jcsd
  3. Nov 20, 2006 #2

    OlderDan

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    There is more information in the graph than you are seeing. What is going on before t = 0? The impulse is the area under the curve of the graph of the net force vs time.
     
  4. Nov 20, 2006 #3
    ohhh, so you're saying that the impulse is actually 0.4N * 1.5s = 0.6Ns? So actually the equation is 0.6 = -mvi. So now what do I do next?
     
  5. Nov 20, 2006 #4
    You can find the mass of the athelete from the graph before 0 sec. Then, plug and solve for v_i.
     
  6. Nov 20, 2006 #5
    Hmm... I don't see what ur saying. How can u get Vi from the graph before 0 sec. Am I missing some concept here?
     
  7. Nov 20, 2006 #6
    w=gm

    Weight is in N, g is gravity, m is in kg.
     
  8. Nov 20, 2006 #7
    Alright, I think I'm getting this. So:
    w = mg
    600N/9.8N/kg = 61.2Kg

    Ft = mvf - mvi
    Ft = -mvi
    (1.5s * 400N * .5) * 1.5s = -61.2kg * Vi (in this part, do you have to divide the force by two like I did since the area under the curve is roughly a right triangle?)
    Vi = -7.35 m/s

    Therefore, using the equation x = Vit + 0.5at^2
    x = -7.35 * 1.5 + 0.5 * -9.8 * 1.5^2 = -22.05m

    The answer doesn't seem to be right. I mean at least the sign should be positive
     
  9. Nov 20, 2006 #8

    OlderDan

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    In this equation

    Ft = mvf - mvi

    vi is zero and vf is the velocity when he leaves the floor at the end of the impulse. vf is positive. After that it is a vertical projectile problem. Your Ft is too big. The 400N and .5 parts are good, but the time base of the triangle is not right. Not only that, it seems you have used the time twice. Ft is the area of the triangle. Check the dimensions of your equation. There can be only one time factor on the left.

    You might want to use the equation that relates the change in velocity squared to the acceleration and the distance for the flight (or conservation of energy which is the same thing).
     
  10. Nov 21, 2006 #9
    okay, I think I kinda know what ur saying. I'll give it an another shot
    Ft = mvf
    400N * 1.0s * 0.5 = 61.2 kg * vf
    vf = 3.26 m/s

    x = vit + 0.5at^2
    x = 3.26 * 1s + 0.5*9.8*1^2
    = 8.16 m

    Is what I am doing correct? I don't get what u said about "You might want to use the equation that relates the change in velocity squared to the acceleration and the distance for the flight (or conservation of energy which is the same thing)." Sorry about this, but I am really bad on this Impulse thing (I was doing perfectly fine on the other conservation of mometum thingy)
     
  11. Nov 21, 2006 #10

    OlderDan

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    I still think your time for the implulse calculation is a bit too long. If you extend the sloping line down to the .6kN line it intersects at about .3sec, so the base of the triangle is about .7sec. The time in the air has no direct connection to the impulse time. It depends only on the velocity achieved at the end of the impulse. The equation that will tell you the maximum height is

    v² = v_o² + 2as

    At maximum height the velocity is zero, the initial velocity is what you caclulate at the end of the impulse, and a is -g. This equation is equivalent to energy conservation

    ½mv² + mgh = ½mv_o²
     
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