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Terminal velocity help

  1. Sep 20, 2005 #1
    Hi. I could use some help from someone who is pretty good at physics.

    Me and a few others are doing a project (that has nothing to do with physics) and we need to know the velocity of a person at the time of impact who fell from 100 meters. We figured out that it would be about 100miles per hour/44meters per second without air resistance, but we need to know what it would be WITH air resistance. This person is about 5'10'' and weighs about 150lbs/68.04kg. It's for a sorta mock trial. If anyone could help, we'de really appreciate it!
     
  2. jcsd
  3. Oct 2, 2005 #2

    hotvette

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    At that short of a distance, I doubt that air drag has much of any effect. Basically, air drag is proportional to the square of velocity. The proportionality constant (called drag coefficient) is dependent mainly on the shape of the object. Example - an arrow has minimal air drag compared to a sphere of the same volume.

    Anyway, if you want to dig into it, Google searches with terms like 'terminal velocity', 'air drag', 'drag coefficient', will probably get you everything you wanted to know.
     
  4. Oct 2, 2005 #3

    arildno

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    Assuming you jump off from rest, and that air resistance is modelled as proportional to the square of the velocity, it is relatively easy to derive the velocity dependence on h, the distance fallen from the jump off point:
    [tex]v(h)=\frac{mg}{k}\sqrt{1-e^{-\frac{2kh}{m}}}[/tex]
    where m is the object's mass, k the air resistance proportionality constant.
    Note that in the limit [itex]\frac{kh}{m}\to{0}[/tex], we get [itex]v(h)=\sqrt{2gh}[/tex] as we of course should have.
    (This include the special cases of zero air resistance or infinite mass of the falling object)
     
    Last edited: Oct 2, 2005
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