Final velocity of a falling object

[pumpui]

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 $v_{1}$. If the value $g$ 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 $v_{1}+v_{1}$ or $2v_{1}$."

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

Here's my thinking:

From an equation, $v^{2}_{f}=v^{2}_{i}+2gs$, then we have
$v^{2}_{1}=2g(1000)$ and $v^{2}_{2}=2g(2000)$, and thus
$v_{2}=\sqrt{2}v_{1}$.

 

[BvU]

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.

 

[pumpui]

Thank you!

 

[pikpobedy]

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.

 

[sophiecentaur]

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.