Time Taken for Body to Fall at Earth's Surface
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Theorem
Let an object $m$ be released above ground from a point near the Earth's surface and allowed to fall freely.
Let $m$ fall a distance $s$ in time $t$.
Then:
- $s = \dfrac 1 2 g t^2$
or:
- $t = \sqrt {\dfrac {2 s} g}$
where $g$ is the Acceleration Due to Gravity at the height through which $m$ falls.
It is supposed that the distance $s$ is small enough that $g$ can be considered constant throughout.
Proof
From Body under Constant Acceleration: Distance after Time:
- $\mathbf s = \mathbf u t + \dfrac {\mathbf a t^2} 2$
Here the body falls from rest, so:
- $\mathbf u = \mathbf 0$
Thus:
- $\mathbf s = \dfrac {\mathbf g t^2} 2$
and so taking magnitudes:
- $s = \dfrac {g t^2} 2$
It follows by multiplying by $\dfrac 2 g$ that:
- $t^2 = \dfrac {2 s} g$
whence:
- $t = \sqrt {\dfrac {2 s} g}$
$\blacksquare$
Sources
- 2008: Ian Stewart: Taming the Infinite ... (previous) ... (next): Chapter $4$: Lure of the Unknown: The logic of species