# Final velocity of a falling object

by pumpui
Tags: falling, final, object, velocity
 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 $v_{1}$. 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 $v_{2}=\sqrt{2}v_{1}$.