How to Derive Free Fall Model with Simple Resistance (Velocity-Time)?

General Solution

Given the differential equation, $\frac{dv}{dt}+\frac{k}{m}v=g$, find its general solution:

Let $A=\frac{k}{m}$:

$\frac{dv}{dt}+Av=g$

Apply the separation of variables and integrate both sides of the equation:

$dv=(g-Av)dt$

$\int \frac{dv}{g-Av}=\int dt$

$-\frac{1}{A}\ln (g-Av)=t+C_1$

Distribute $A$:

$\ln (g-Av)=-At-C_{1}A$

Convert to exponential notation:

$g-Av=e^{-At}{e}^{-C_{1}A}$

Let $e^{-C_{1}A}=C$:

$g-Av=C{e}^{-At}$

Isolate $v$:

$v=\frac{g-C{e}^{-At}}{A}$

Return $A$:

$v=\frac{m(g-C{e}^{-\frac{k}{m}t})}{k}$

Particular Solution

Let's find a particular solution to model the event of free-fall with simple resistance. In free-fall, we know that the initial velocity is zero at the very start $t=0$. If we substitute this into the general solution, we will obtain $C=g$; hence, the particular answer is:

$v=\frac{mg(1-e^{-\frac{k}{m}t})}{k}$