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

General Solution

Given the differential equation, \(\frac{d v}{d t}+\frac{k}{m} v=g\), find its general solution:

Let \(A=\frac{k}{m}\):

\(\frac{d v}{d t}+A v=g\)

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

\(d v=(g-A v) d t\)

\(\int \frac{d v}{g-A v}=\int d t\)

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

Distribute \(A\):

\(\ln (g-A v)=-A t-C_1 A\)

Convert to exponential notation:

\(g-A v=e^{-A t} e^{-C_1 A}\)

Let \(e^{-C_1 A}=C\):

\(g-A v=C e^{-A t}\)

Isolate \(v\):

\(v=\frac{g-C e^{-A t}}{A}\)

Return \(A\):

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

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:

\(v=\frac{m g\left(1-e^{-\frac{k}{m} t}\right)}{k}\)