Speed of Body under Free Fall from Height/Proof 1

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:
 * $v = \sqrt {2 g s}$