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Final velocity of a falling object

by pumpui
Tags: falling, final, object, velocity
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Jan20-14, 05:15 PM
P: 2
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
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Jan20-14, 06:47 PM
HW Helper
P: 1,373
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.
Jan20-14, 08:11 PM
P: 2
Thank you!

Jan20-14, 08:39 PM
P: 39
Final velocity of a falling object

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.
Jan21-14, 04:16 AM
Sci Advisor
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sophiecentaur's Avatar
P: 12,186
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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