Let's explore free-fall motion with fluid resistance in a real-life scenario and how such motion models help us understand the event.
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Continuing our previous post, we have created motion models for a free-falling object with air resistance. For fluid resistance \(F_f=kv^2\), these are:

  • \(v(t)=\sqrt{\frac{m g}{k}} \tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\)
  • \(a(t)=-g \operatorname{sech}^2\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\)
  • \(s(t)=s_0-\frac{m \ln \left(\cosh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\right)}{k}\)

With these models, we can describe the motion of a free-falling object with drag resistance. Now, let's see how we can exactly apply these models.


Let's say we're designing parachutes. Imagine we're making these life-saving devices. Before we can make one, we need to know what is happening with the falling object.

Say we have an object that fell from the top of a building:

  • The falling thing has a mass of 65 kg \(m\)
  • We drop it \(v_0=0\) from a height of 800m, initial position \(s_0\), from the ground floor.
  • Gravitational acceleration is 9.81 \(\frac{m}{s^2}\)
  • Variable \(k\) is equal to 0.035
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From there, we can now describe the motion of this instance. Using these values, we substitute them into the three equations. Below are the corresponding motion graphs of these equations:




We can interact with these graphs to determine the position, velocity, and acceleration of the falling body at any time. Note that velocity is negative, which refers to the object's direction.

Some notable observations include the following:

  • From the position-time graph, it will only take 13.705 seconds for the body to reach the ground from a height of 800 m. If you're designing a parachute for this instance, then it means the parachute has to deploy faster than that.
  • When it has reached the ground, the body has a final velocity of around 103 m/s (371 kph) and a deceleration of about 4.1 m per second squared. That is extremely fast compared to typical speed limits you encounter on expressways. What's more concerning is that it is a fatal crash speed. In parachute design, you must reduce the final impact speed so the user is safe when they reach the ground. That means that designers will need to find a way to make deceleration constant so that it doesn't gain speed, plus the parachute can slow down the fall.

From this application, we can see how we can use these functions to study the behavior of the free-falling motion with air resistance.

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

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