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Terminal velocity in finite time

  1. Apr 24, 2006 #1
    [tex]v^2 = \frac{mg}{k}(1 - e^{-2ky/m})[/tex]

    As [itex]t \rightarrow \infty[/itex] and [itex]y \rightarrow \infty[/itex] we see [itex]TV = \sqrt{mg/k}[/itex]

    And below my book reads: "From actual experience we know that a raindrop reaches its limiting velocity in a finite and not an infinite amount of time. This is because other factors also operate to slow the raindrop's velocity."

    What are these factors? o:)
    Last edited: Apr 24, 2006
  2. jcsd
  3. Apr 24, 2006 #2
    I don't know what he would mean by this. The drag force is what's slowing down the raindrop, so friction can't be an answer. K depends on the geometry of the raindrop and the fluid. The K should change a small amount as the density of the fluid increases as the raindrop gets lower. But appart from that, I don't know. Maybe Clausius can tell us why.
  4. Apr 24, 2006 #3


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    Maybe the passage means that the fluctuations in the value of k as the object falls change the velocity more than the difference between v and vt
  5. Apr 25, 2006 #4
    I think you're right because before he states some approximations so that air resistance is R = kv.
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