Final velocity of a falling object

  1. Jan 20, 2014 #1
    I was reading The History of Physics by Isaac Asimov, and I came across this passage.

    "Imagine a body dropped first from a height of 1000 kilometers, then from 2000 kilometers, then from 3000 kilometers, and so on. The drop from 1000 kilometers would result in a velocity of impact [itex]v_{1}[/itex]. If the value [itex]g[/itex] were constant all the way up, then a drop from 2000 kilometers would involve a gain in the first 1000 kilometers equal to the gain in the second 1000 kilometers, so the final velocity of impact would be [itex]v_{1}+v_{1}[/itex] or [itex]2v_{1}[/itex]."

    I was wondering why it came to [itex]2v_{1}[/itex]. Wouldn't it be [itex]\sqrt{2}v_{1}[/itex]?

    Here's my thinking:

    From an equation, [itex]v^{2}_{f}=v^{2}_{i}+2gs[/itex], then we have
    [itex]v^{2}_{1}=2g(1000)[/itex] and [itex]v^{2}_{2}=2g(2000)[/itex], and thus
    [itex]v_{2}=\sqrt{2}v_{1}[/itex].
     
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  3. Jan 20, 2014 #2

    BvU

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    You are quite right. The writer lures you into thinking the time needed to traverse the second 1000 m is equal to the time needed for the first. It definitely is not, because the initial speed for the first is 0 but for the second it is v1.
     
  4. Jan 20, 2014 #3
    Thank you!
     
  5. Jan 20, 2014 #4
    Under constant acceleration the V increases linearly with time. V does not increase linearly with distance.

    This looks correct:
    If we let i = initial height,
    the current height H = i - (a * t^2 / 2).
    So a * t^2 = 2 * (i - H).
    Therefore t = sqrt(2 * (i - H) / a).
    If we substitute that into v = a*t,
    v = a * sqrt (2* (i-H)/a). Note that i and a are constants, so the only independent variable is the current height H.
     
  6. Jan 21, 2014 #5

    sophiecentaur

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    In a nutshell: Kinetic Energy will increase linearly with distance (uniform g) because Potential Energy is reducing linearly (mgh). So the speed will increase as the square root of distance fallen.
     
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