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Impulse-Momentum Theorem and diving board

  1. Jan 27, 2009 #1
    1. The problem statement, all variables and given/known data
    An 82 kg man drops from rest on a diving board 3m above the surface of the water and comes to rest 0.55 s after reaching the water. What force does the water exert on him?

    2. Relevant equations
    F = mVf - mVi / t

    3. The attempt at a solution
    I have no idea where to begin since I don't know the diver's velocity. Could I possibly use V = d/t?

    EDIT: Ok, so I used V = d/t and got 5.5 m/s for the velocity. Then I used F = mVf - mVi / t, and got 820 N. Is this correct?
  2. jcsd
  3. Jan 27, 2009 #2
    I'm not saying that I'm in any way a physics expert or anything but;

    What you calculated as the velocity was 3m/0.55s --> which is not at all the velocity of the diver just before he hits the water. I made a quick attempt at this problem, and I would instead use the relationship: x=at2/2, since the acceleration is constant. From this you can find the time it takes for the diver to hit the water--also knowing that x=vt, and therefore v=x/t, you can find the velocity of the diver just before he hits the water, which is also the initial velocity in (mVf-mVi)/t.
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