# 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 $v_{2}=\sqrt{2}v_{1}$.
 Thanks P: 1,019 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.
 P: 2 Thank you!
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.
 Sci Advisor PF Gold P: 11,382 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.

 Related Discussions Introductory Physics Homework 6 Introductory Physics Homework 3 Calculus & Beyond Homework 13 General Physics 15 Calculus & Beyond Homework 8