The calculations for free-falling with air resistance are slightly more complex than the ones we use for a vacuum. Follow our short article to learn why. From an introduction to the force of air resistance to the formulas, we will learn how to calculate the air resistance on a falling object, the parameters of the fall, and the possible results. To conclude, we will talk about squirrels.

Not-so-free-falling: plunging in a gravitational field, but with air: free-fall with air resistance

Physicists like to analyze a problem starting from a very idealized version of the real world. No outside forces, no interactions: studying this situation is much more manageable, and you can add more "realistic" elements.

The same applies to falling in a gravitational field. We already analyzed in detail the situation where air resistance is negligible in our free fall calculator.

How does the situation change if we include air resistance in our formula to calculate the dynamics of a falling body? Well, from a general point of view, two factors play into a fall with the effects of air resistance (or any other fluid): the force of friction and turbulence. However, studying fluids is not easy: follow us to discover the free-fall with air resistance equations and explanations.

Calculating free fall with air resistance: formula for the force of friction

The first thing we can calculate is the force of air resistance: its formula is straightforward but with a twist.

F=kv2F = k\cdot v^2

Where:

  • FF — The force of air resistance;
  • kk — The air resistance coefficient; and
  • vv — The instantaneous speed.

As you can see, the force depends on the square of the speed: the increase is not linear, but the effect of air resistance is felt more heavily at higher speeds.

The equation for the air resistance coefficient

The value of the air resistance coefficient is of fundamental importance in the calculations of free fall with air resistance. The value of kk contains all the information related to the body and the fluid. Here is its formula:

k=ρAC2k = \frac{\rho\cdot A \cdot C}{ 2}

Where:

  • ρ\rho — The density of the fluid where we are calculating the free fall with resistance: air, water, methane, etc.;
  • AA — The cross-sectional area of the falling body; and
  • CC — A dimensionless drag coefficient related to the shape of the body.

🙋 For a deeper analysis of the value of the drag coefficient, visit our drag equation calculator.

How to calculate air resistance on a falling object: maximum and terminal speed

Once you have the result of the force of air resistance formula, you can make a first qualitative observation. If the value is greater or equal to the one of the gravitational force, the body has reached its terminal velocity, the maximum velocity allowed in that fluid. We discussed this phenomenon in greater detail at our terminal velocity calculator.

To find the terminal velocity vtv_{\text{t}}, equate the two formulas (for air resistance and gravitational force):

mg=kvt2m\cdot g = k\cdot v_{\text{t}}^2

And isolate the value of the speed:

vt=mgk v_{\text{t}}= \sqrt{\frac{m\cdot g}{k}}

If the body doesn't reach the terminal velocity, we can calculate the maximum velocity. To do so, calculate the net force, and then we set up a differential equation. The solution is:

v=mgktanh(tgkm)v = \sqrt{\frac{m\cdot g}{k}}\cdot \tanh \left(t\cdot \sqrt{\frac{g\cdot k}{m}}\right)

tt is the time of fall. We calculate it with the formula:

t=mgkacosh(ehkm)t = \sqrt{\frac{m\cdot g}{k}}\cdot \text{acosh}\left(e^{\frac{h\cdot k}{m}}\right)

Where acosh(x)\text{acosh}(x) is the inverse hyperbolic cosine function, and hh the height of the fall.

As you can see, both speeds (terminal and maximum) depend on the mass of the falling object. We are finally out of the ideal free-fall regime, where the mass doesn't affect the velocity but only the force, and the behavior modeled by our free-fall with air resistance equations better reflects the one we can observe in real life. Let's drop something now!

Dropping squirrels: application of the free-fall with air resistance formulas

We like squirrels, and we like how they behave when falling. They are, in fact, one of the only mammals able to survive a fall from any height (we are not sure about bats). How so? They are fluffy. Tremendously fluffy. All their fur contributes to increasing the drag force, and they can achieve a non-lethal terminal speed.

Let's find the parameter for a squirrel. For the drag coefficient, we will use the one of a skydiver; it should be fine: CD=0.24 kg/mC_{\text{D}} = 0.24\ \text{kg}/\text{m}. The mass is m=0.450 kgm = 0.450\ \text{kg}, our squirrel is a fit one. The critter got drunk after eating too many fermented crab apples, and lost his balance from a h=2.5 mh = 2.5\ \text{m} tree. At which fall will he reach the ground?

v=mgktanh(tgkm)=v=0.459,810.24tanh(0.889,810.240.45)=4.14 ms=14.9 kmh\begin{align*} v &= \sqrt{\frac{m\cdot g}{k}}\cdot \tanh \left(t\cdot \sqrt{\frac{g\cdot k}{m}}\right)\\ \\ &=v = \sqrt{\frac{0.45\cdot 9,81}{0.24}}\\ \\ &\cdot \tanh \left(0.88\cdot \sqrt{\frac{9,81\cdot 0.24}{0.45}}\right)\\ & =4.14\ \frac{\text{m}}{\text{s}} = 14.9\ \frac{\text{km}}{\text{h}} \end{align*}

Our squirrel will be fine. Now let's climb on an SR-71, a plane able to cruise at 26 km26\ \text{km} of altitude. Give a pressure suit to our almost squirrelnaut, and wish him good luck. Jump!

We will reach terminal speed, this time: let's find it.

vt=mgk=0.459,810.24=4.29 ms=15.44 kmh \begin{align*} v_{\text{t}}&= \sqrt{\frac{m\cdot g}{k}} = \sqrt{\frac{0.45\cdot 9,81}{0.24}}\\ \\ &= 4.29\ \frac{\text{m}}{\text{s}} = 15.44\ \frac{\text{km}}{\text{h}} \end{align*}

It's barely more than the speed we found after the jump from the tree! Squirrels are awesome!

Davide Borchia
Traveling object
Mass (m)
kg
Altitude (h)
m
Gravitational acceleration (g)
m/s²
Air resistance coefficient (k)
kg/m
Time and velocity
Time of fall (t)
sec
Maximum velocity (vmax)
m/s
Terminal velocity (vt)
m/s
Drag force
at velocity...
m/s
Drag force is
N
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