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:


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