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Force exerted by water on diver (2nd law of Motion)

  1. Oct 21, 2009 #1
    I've been in my physics class for almost two months now, and I'm starting to run into trouble. I think I'll be around here for a while, assuming you guys can help me out ;)

    The Question : A high diver of mass 68.4 kg jumps off a board 9.5 m above the water.
    The acceleration of gravity is 9.8 m/s2 .
    If his downward motion is stopped 5.92 s after he enters the water, what average upward
    force did the water exert on him?
    Answer in units of N.

    2. Relevant equations :
    v = v0 + at (To determine acceleration
    EF (Average Force) = ma

    3. The attempt at a solution :
    Well, I first have to find the acceleration of the water on the swimmer :
    v = v0 + at
    0 = (9.5m*9.8m/s^2) + a (5.92)
    -93.1 = a (5.92s)
    -15.726 m/s^2 = a

    Now I plug that into F = ma
    F = ma = (68.4 kg) (-15.726 m/s^2) = -1075.682 N

    This looks about right, but when I put it into Quest, it comes out wrong. I've checked my math over twice. What am I doing wrong?

    Thanks for the help :)
     
  2. jcsd
  3. Oct 21, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    you put t=5.92, but that is for when he now enters the water.

    so in v=u+at

    we want v=0, but the diver doesn't enter the water with 0 velocity. So you need to find this velocity 'u'. (try using conservation of energy to find this velocity)
     
  4. Oct 22, 2009 #3
    Thank you very much - that helped me a bunch ;)
     
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