Final velocity of a falling object

  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
  2. jcsd
  3. BvU

    BvU 4,837
    Homework Helper
    Gold Member
    2014 Award

    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. Thank you!
  5. 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. sophiecentaur

    sophiecentaur 14,696
    Science Advisor
    Gold Member

    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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