Speed of Body under Free Fall from Height/Proof 1

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Theorem

Let an object $B$ be released above ground from a point near the Earth's surface and allowed to fall freely.

Let $B$ fall a distance $s$.

Then:

$v = \sqrt {2 g s}$

where:

$v$ is the speed of $B$ after having fallen a distance $s$
$g$ is the Acceleration Due to Gravity at the height through which $B$ falls.

It is supposed that the distance $s$ is small enough that $g$ can be considered constant throughout.


Proof

From Body under Constant Acceleration: Velocity after Distance:

$\mathbf v \cdot \mathbf v = \mathbf u \cdot \mathbf u + 2 \mathbf g \cdot \mathbf s$

All dot products are between pairs of parallel vectors.

Thus by Cosine Formula for Dot Product:

$v^2 = u^2 + 2 g s$

Here the body falls from rest, so:

$\mathbf u = \mathbf 0$

Thus:

$v^2 = 2 g s$

and so:

$v = \sqrt {2 g s}$

$\blacksquare$