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Freefall Acceleration of a Raindrop

  1. Sep 15, 2004 #1
    I have an answer that does not seem right:

    The raindrop falls from 1700m high with constant acceleration of 9.81 m / s^2
    What is the velocity of the raindrop not considering any resistance?

    i used y - yo = vo * t - (1/2) g t^2 to get the time to hit the ground at 18.62 seconds.

    I then plugged that into v = vo + a t to get 182.63 m / s which seems kind of fast for a raindrop, right? Can anyone help? Please?
  2. jcsd
  3. Sep 15, 2004 #2


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    So that would imply, that in reality, you cannot ignore air resistance, if you want a meaningful value for the velocity of a rain drop. I did not check your numbers but they do not seem unreasonable, for this ideal but unrealistic scenario.

    Edit, checked your numbers they are right in all of the signifiant digits.
    Last edited: Sep 15, 2004
  4. Sep 15, 2004 #3
    Thank you for the help :approve: I always get worried that I did something wrong when numbers do not turn out to appear reasonable.
  5. Sep 16, 2004 #4


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    Check out Stokes' resistance law on a sphere moving through a viscous fluid.
    This yields a fall velocity consistent with experiments on the fall velocity of fog droplets.
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