Speed of Body under Free Fall from Height

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 1

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:

$v = \sqrt {2 g s}$

$\blacksquare$

Proof 2

From Acceleration is Second Derivative of Displacement with respect to Time:

$\mathbf g = \dfrac {\d^2 \mathbf s} {\d t^2}$

Integrating with respect to $t$, and by definition of velocity:

$\mathbf v = \dfrac {\d \mathbf s} {\d t} = \mathbf g t + \mathbf c_1$

When $t = 0$, we have that $\mathrm c_1$ is the initial velocity $\mathbf v_0$, and so:

$\mathbf v = \dfrac {\d \mathbf s} {\d t} = \mathbf g t + \mathbf v_0$

Integrating with respect to $t$ again:

$\mathbf s = \dfrac {\mathbf g t^2} 2 + \mathbf v_0 t + \mathbf c_2$

When $t = 0$, we have that $\mathrm c_2$ is the initial displacement $\mathbf s_0$, and so:

$\mathbf s = \dfrac {\mathbf g t^2} 2 + \mathbf v_0 t + \mathbf s_0$

We have that $B$ is released at rest starting at $\mathbf s = \mathbf 0$.

Thus $\mathbf v_0 = \mathbf s_0 = \mathbf 0$ and so:

$\mathbf s = \dfrac {\mathbf g t^2} 2$

$\mathbf v = \mathbf g t$

By eliminating $t$ and taking the scalar quantities:

$v = \sqrt {2 g s}$

$\blacksquare$

Proof 3

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.

$\blacksquare$