Speed of Body under Free Fall from Height/Proof 2
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
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems