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:
- $v = \sqrt {2 g s}$
$\blacksquare$