# Speed of Body under Free Fall from Height/Proof 3

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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 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.

$\blacksquare$

## Sources

- 1972: George F. Simmons:
*Differential Equations*... (previous) ... (next): $\S 1.5$: Falling Bodies and Other Rate Problems