Typically in ideal free-fall motion, we study an object's movement with air resistance neglected. It means the body encounters no friction while it falls – like a body falling in a vacuum.

In real life, most objects will have some fluid resistance force that counteracts the motion of an object. This post explores that phenomenon.

## Modeling Free Fall Motion with Air Resistance

Let's begin by setting up the situation – an object falls into Earth's atmosphere due to gravity. While it is dropping, air resistance counters its motion. We want to model the movement of the object due to these factors.

We describe its motion with the three critical motion variables against time \(t\) from a frame reference: position \(s\), velocity \(v\), and acceleration \(a\).

At the end of this post, we'll have three different models to describe the body's motion:

- \(s-t\), position-time
- \(v-t\), velocity-time
- \(a-t\), acceleration-time

## Setting-Up The Model

Let's study the object's kinetics and use Newton's Second Law of Motion. Using its elementary definition, the acceleration \(a\) of an object is directly proportional to the magnitude of the net force \(F_n\) and inversely proportional to the mass \(m\):

\(a=\frac{F_n}{m}=\frac{\sum F_y}{m} \Longrightarrow \sum F_y=m a\)

We identify the forces acting on a small falling object of mass \(m\).

- The force mainly responsible for the fall, the gravitational force between the falling object and Earth, \(F_g\) (or weight)
- The force resisting the motion, which is fluid resistance \(F_f\)

Consequently, Newton's second law would then result in the following equation:

- \(\left.\sum F_y=m a\right] \uparrow_{+}\)
- \(F_f+\left(-F_g\right)=m a\)
- \(F_f-F_g=m a\)

Going sidetrack for sign convention, we follow the grid defined in the figure:

- The fluid resistance is positive because it goes upwards, while the gravitational force is negative because it goes downwards.
- In other references, the direction of the positive sense for free fall motion is downwards (hence, the gravitational force is positive while the resistance is negative).

We can choose whatever sign convention as long as it's clearly defined.

Now that we know the different factors affecting our model, we proceed to derive to find these models: \(v-t\), \(a-t\), and \(s-t\). To start, we recall that:

- \(F_f-F_g=m a\)
- \(F_f-m g=m a\)

At this point, we can express fluid resistance in two ways. We will consider each case: \(F_f=kv\) and \(F_f=kv^2\). It implies that free fall motion with air resistance can have two variations:

### Fluid Resistance \(F_f=kv\)

Let's consider the case when \(F_f=kv\):

- \(F_f-m g=m a\)
- \(k v-m g=m a\)

From any Calculus or Physics reference, instantaneous acceleration is:

- \(a(t)=\lim _{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t}=\frac{d v}{d t}\)

Substituting this to our primary expression, we can obtain a differential equation:

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

Rearranging to standard form:

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

The previous expression is the primary equation we will use to describe the model of the falling object (in terms of velocity-time).

Solving the differential equation deserves its post, which you can check out here.

#### Model

After solving, we will end up with the following general solution, which is the velocity-time model:

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

After finding the velocity-time function, we can solve for acceleration-time and position-time using key motion relationships:

\(a(t)=g e^{-\frac{k}{m}t}\)

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

- \(v(t)\) is the velocity of the falling body as a function of time
- \(a(t)\) is the acceleration of the falling body as a function of time
- \(s(t)\) is the position of the falling body as a function of time
- \(m\) is the mass of the object
- \(g\) is the acceleration due to gravity
- \(k\) is the proportionality constant (from fluid resistance)
- \(t\) is a specific time in motion

### Fluid Resistance \(F_f=kv^2\)

Let's consider the case when \(F_f=kv^2\):

- \(F_f-m g=m a\)
- \(k v^2-m g=m a\)

From any Calculus or Physics reference, instantaneous acceleration is:

- \(a(t)=\lim _{\Delta t \rightarrow 0} \frac{\Delta v}{\Delta t}=\frac{d v}{d t}\)

We substitute this to our model to obtain a first-order ordinary differential equation:

- \(k v^2-m g=m \frac{d v}{d t}\)
- \(\frac{k v^2}{m}-g=\frac{d v}{d t}\)

For simplicity in the solution, let's make \(\frac{k}{m}=a\) first:

- \(a v^2-g=\frac{d v}{d t}\)

Applying the separation of variables method, we get the primary equation that we will use to figure out the model:

- \(\frac{1}{a v^2-g} d v=d t\)

The previous expression is the primary equation we will use to describe the model of the falling object (in terms of velocity-time).

Solving the differential equation deserves its post, which you can check out here.

#### Model

After solving, we will end up with the following general solution, which is the velocity-time model:

\(v(t)=\sqrt{\frac{m g}{k}} \tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\)

After finding the velocity-time function, we can solve for acceleration-time and position-time using key motion relationships:

\(a(t)=-g \operatorname{sech}^2\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\)

\(s(t)=s_0-\frac{m \ln \left(\cosh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\right)}{k}\)

- \(v(t)\) is the velocity of the falling body as a function of time
- \(a(t)\) is the acceleration of the falling body as a function of time
- \(s(t)\) is the position of the falling body as a function of time
- \(m\) is the mass of the object
- \(g\) is the acceleration due to gravity
- \(k\) is the proportionality constant (from fluid resistance - drag)
- \(t\) is a specific time in motion
- \(v_0\) is the initial velocity
- \(s_0\) is the initial position

## Summary

Typically in ideal free-fall motion, we study an object's movement with air resistance neglected; however, in real life, most things will have some resistive force that counteracts its movement, which we usually call fluid resistance.

When an object falls into Earth's atmosphere due to gravity, it experiences two forces: gravitation \(F_g\) and fluid resistance \(F_f\) force.

Gravitational force \(F_g\) is due to the product of mass \(m\) and acceleration due to gravity \(g\).

We can express fluid resistance \(F_f\) in two ways: \(F_f=kv\) and \(F_f=kv^2\).

In \(F_f=kv\), the free-fall motion models are: \(v(t)=\frac{m g\left(1-e^{-\frac{k}{m} t}\right)}{k}\), \(a(t)=g e^{-\frac{k}{m}t}\), and \(s(t)=\frac{m g}{k}\left(t+\frac{m}{k} e^{-\frac{k}{m} t}\right)\)

In \(F_f=kv^2\), the free-fall motion models are: \(v(t)=\sqrt{\frac{m g}{k}} \tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\), \(a(t)=-g \operatorname{sech}^2\left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\), and \(s(t)=s_0-\frac{m \ln \left(\cosh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\right)}{k}\)

\(v(t)\) is the velocity of the falling body as a function of time, \(a(t)\) is the acceleration of the falling body as a function of time, \(s(t)\) is the position of the falling body as a function of time, \(m\) is the mass of the object, \(g\) is the acceleration due to gravity, \(k\) is the proportionality constant (from fluid resistance - drag), \(t\) is a specific time in motion, \(v_0\) is the initial velocity, and \(s_0\) is the initial position