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Final velocity of a falling object 
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#1
Jan2014, 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 [itex]v_{2}=\sqrt{2}v_{1}[/itex]. 


#2
Jan2014, 06:47 PM

HW Helper
Thanks
P: 1,352

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.



#3
Jan2014, 08:11 PM

P: 2

Thank you!



#4
Jan2014, 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* (iH)/a). Note that i and a are constants, so the only independent variable is the current height H. 


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