I was reading The History of Physics by Isaac Asimov, and I came across this passage.(adsbygoogle = window.adsbygoogle || []).push({});

"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].

**Physics Forums - The Fusion of Science and Community**

# Final velocity of a falling object

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Final velocity of a falling object

Loading...

**Physics Forums - The Fusion of Science and Community**