Motion of Body Falling through Air

Theorem

The motion of a body $B$ falling through air can be described using the following differential equation:

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

where:

$m$ denotes mass of $B$

$y$ denotes the height of $B$ from an arbitrary reference

$t$ denotes time elapsed from an arbitrary reference

$g$ denotes the local gravitational constant acting on $B$

$k$ denotes the coefficient of resistive force exerted on $B$ by the air (assumed to be proportional to the speed of $B$)

Proof

From Newton's Second Law of Motion, the force on $B$ equals its mass multiplied by its acceleration.

Thus the force $F$ on $B$ is given by:

$F = m \dfrac {\d^2 y} {\d t^2}$

where it is assumed that the acceleration is in a downward direction.

The force on $B$ due to gravity is $m g$.

The force on $B$ due to the air it is passing through is $k$ multiplied by the speed of $B$, in the opposite direction to its travel.

That is::

$k \dfrac {d y} {d t}$

Hence the required differential equation:

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

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $1$: The Nature of Differential Equations: $\S 1$: Introduction: $(3)$