Terminal Velocity of Body under Fall Retarded Proportional to Velocity

Theorem

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

Let $B$ be falling through a medium which exerts a resisting force $k \mathbf v$ upon $B$ which is proportional to the velocity of $B$ relative to the medium.

Then the terminal velocity of $B$ is given by:

$v = \dfrac {g m} k$

Proof

Let $B$ start from rest.

From Motion of Body Falling through Air, the differential equation governing the motion of $B$ is given by:

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

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

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

By definition of velocity:

$\dfrac {\d \mathbf v} {\d t} = \mathbf g - c \mathbf v$

and so:

\(\ds \int \dfrac {\d \mathbf v} {\mathbf g - c \mathbf v}\)

\(=\)

\(\ds \int \rd t\)

\(\ds \leadsto \ \ \)

\(\ds -\dfrac 1 c \map \ln {\mathbf g - c \mathbf v}\)

\(=\)

\(\ds t + c_1\)

\(\ds \leadsto \ \ \)

\(\ds \mathbf g - c \mathbf v\)

\(=\)

\(\ds \mathbf c_2 e^{-c t}\)

When $t = 0$ we have that $\mathbf v = 0$ and so:

$\mathbf c_2 = \mathbf g$

Hence by taking magnitudes:

$v = \dfrac g c \paren {1 - e^{-c t} }$

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

Thus in the limit:

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

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems