Speed of Body under Free Fall from Height/Proof 3

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): $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems