Speed of Body under Free Fall from Height/Proof 3

Proof
From the Principle of Conservation of Energy:
 * $K + P = C$

where:
 * $K$ is the kinetic energy of $B$
 * $P$ is the potential energy of $B$
 * $C$ is a constant.

Let the mass of $B$ be $m$.

From Kinetic Energy of Motion:
 * $K = \dfrac {m v^2} 2$

where $v$ is the speed of $B$.

From Potential Energy of Position:
 * $P = m g s$

where $s$ is the distance fallen by $B$.

Since $B$ falls from rest, its initial kinetic energy is zero.

Having fallen a distance $s$, $B$ has lost potential energy $m g s$.

Therefore:
 * $\dfrac {m v^2} 2 = m g s$

from which the result follows by dividing both sides by $m$ and extracting the square root.