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Drag forces on someone diving into a pool

  1. Sep 30, 2008 #1
    1. The problem statement, all variables and given/known data

    You dive straight down into a pool of water. You hit the water with a speed of 7.0 m/s, and your mass is 75 kg.

    Assuming a drag force of the form FD= (−1.00×104 kg/s)v, how long does it take you to reach 2% of your original speed? (Ignore any effects of buoyancy.)

    2. Relevant equations

    [tex]\Sigma[/tex]F=ma
    V=Vo+at (?)

    3. The attempt at a solution

    I summed the forces in the y-direction (down is positive):
    [tex]\Sigma[/tex]F=mg-FD=may

    2% of 7 m/s = 0.14 m/s

    But I don't know what to do from here. How do I solve for ay when the drag force varies with velocity? If ay varies, then how can I use the kinematic equations for constant a?

    I figured this out the hard way after plugging in 7.0 m/s for v and solving for a, which is an absurdly large number. (-923 m/s2)
     
  2. jcsd
  3. Sep 30, 2008 #2
    Are you sure that FD=(-1.00x104)v?
    This coefficient seems kind of large to be for drag, recheck the numbers.
     
    Last edited: Sep 30, 2008
  4. Sep 30, 2008 #3
    Lazyshot,

    I should've placed the negative sign outside the parenthesis, but yes, according to my book and MasteringPhysics, that is the coefficient.

    FD=-(1.00x104)v
     
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