Speed of Body under Free Fall from Height/Proof 2

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:
 * $v = \sqrt {2 g s}$