Speed of Body under Free Fall from Height/Proof 2

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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 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:

$v = \sqrt {2 g s}$

$\blacksquare$


Sources