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Skydiver drag problem

  1. Sep 25, 2006 #1


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    A 75.0 kg skydiver can be modeled as a rectangular "box" with dimensions 24.0 cm\times 44.0 cm\times 184 cm.
    What is his terminal speed if he falls feet first?

    I know that drag = mg when terminal speed is reached. so drag is 75*9.8

    and there's a formula drag = KV^2 where K is drag coeff and V is velocity, but I don't know K.

    a little help?
  2. jcsd
  3. Sep 25, 2006 #2
    Okay, so you summed the forces when they were in equilibrium and found the force of drag. You don't know K, does that mean you are supposed to derive it? If so, my hint is that you were given two different quantities in your problem, and you already used one... so K must have something to do with the other.
  4. Sep 26, 2006 #3


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    [tex]K = \frac{1}{2} \rho S C_D[/tex]
    [tex]\rho[/tex] is air density.
    S is the cross section area. In your case 24cm x 44cm.
    [tex]C_D[/tex] is the drag coefficient, normally determined experimentally. If you don´t have it´s value I don´t see how you could solve the problem.
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