This post shows the derivation of the model between two motion variables, velocity and time, of an object experiencing free-fall motion with simple resistance.
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This post shows the solution of the ordinary differential equation (ODE) for falling bodies with fluid resistance \(F_f = kv\), velocity-time, v-t. If you wish to discover how the ODE was derived, read here first.

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

  • The variable \(v\) refers to the velocity of the falling objects
  • Variable \(t\) refers to the time
  • The variable \(k\) refers to the drag proportionality constant.
  • Variable \(m\) is the object's mass
  • Constant \(g\) refers to the acceleration due to gravity.

Finding the Solution for Velocity-Time

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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:

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

Created On
June 5, 2023
Updated On
February 23, 2024
Contributors
Edgar Christian Dirige
Founder
References

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Revision
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