Terminal Velocity of Body under Fall Retarded Proportional to Square of Velocity

Theorem
Let $B$ be a body falling in a gravitational field.

Let $B$ be falling through a medium which exerts a resisting force of magnitude $k v^2$ upon $B$ which is proportional to the square of the velocity of $B$ relative to the medium.

Then the velocity of $B$ reaches a limiting value:
 * $v = \sqrt {\dfrac {g m} k}$

This value is called the terminal velocity of $B$.

Proof
Let $B$ start from rest.

The differential equation governing the motion of $B$ is given by:


 * $m \dfrac {\mathrm d^2 \mathbf s} {\mathrm d t^2} = m \mathbf g - k \left({\dfrac {\mathrm d \mathbf s} {\mathrm d t} }\right)^2$

Dividing through by $m$ and setting $c = \dfrac k m$ gives:


 * $\dfrac {\mathrm d^2 \mathbf s} {\mathrm d t^2} = \mathbf g - c \left({\dfrac {\mathrm d \mathbf s} {\mathrm d t} }\right)^2$

By definition of velocity:


 * $\dfrac {\mathrm d \mathbf v} {\mathrm d t} = \mathbf g - c \mathbf v^2$

for some constant $c$.

and so taking magnitudes of the vector quantities:

We have $v = 0$ when $t = 0$ which leads to $\ln 1 = 0 + c_1$ and thus $c_1 = 0$.

Hence:

Since $c > 0$ it follows that $v \to \sqrt {\dfrac g c}$ as $t \to \infty$.

Thus in the limit:


 * $v = \sqrt {\dfrac g c} = \sqrt {\dfrac {g m} k}$