When Does the Air Resistance Force Make a Difference?

Rhett Allain

I often look at cases where things are falling. We typically call this “free fall” motion because the object is moving only under the influence of the gravitational force. With only the gravitational force, the object has a constant acceleration and the motion is fairly simple to model.

However, objects on the surface of the Earth usually have an air resistance force on them also. When can we ignore this extra force and when is it important?

Modeling Air Resistance

Let’s say I drop a ping pong ball. As it falls, I can draw the following force diagram.

The most common model for the air resistance says that the magnitude of the force depends on:

• The density of air (ρ). This typically has a value around 1.2 kg/m³.


• The cross sectional area of the object (A). A ping pong ball would have a cross sectional area equal to π*r².


• The drag coefficient (C). This depends on the shape of the object. For a spherical object, a unitless value of 0.47 is typical.


• The magnitude of the velocity squared. The faster you go, the greater the air resistance force.

The direction of the air resistance force is in the opposite direction as the velocity of the object. That’s why there is a negative sign in the expression along with the r – hat (which is a unit vector in the direction of the velocity).

But how do you find values for the drag coefficients for different objects? The real answer is that you must measure them experimentally. However, Wikipedia has a nice list of some values. What about a falling human? I often have to model the motion of a falling human, but there isn’t a C value listed. There is one trick I can use.

The trick involves terminal velocity. Suppose a human jumps out of a stationary hot air balloon. At first, only the gravitational force acts on the human giving an acceleration of -9.8 m/s². However, as the human increases in speed, the air resistance force also increases. At some point, the air resistance force will be equal in magnitude to the gravitational force and the human will no longer increase in speed. We call this “terminal velocity”.

Now for the trick. It seems to be mostly accepted that the terminal velocity for a skydiver is about 120 mph (53.6 m/s). Of course, this is the terminal velocity for the normal skydiving position with head facing down and arms and legs spread out. If I guess at a human mass of 70 kg, I can set the air resistance and gravitational forces equal. Also, for simplicity I am going to call all the constants in front of the velocity squared just K (since they don’t change).

I only need the mass and the terminal velocity and I can build a model for air resistance. Yes, this is just a model. If you go super fast, this model probably isn’t valid. For now, it’s all I have to work with.

How High is Too High?

If I drop an object from some height, there are two things I could do to obtain a value for the falling time. First, I could just ignore air resistance and use the typical kinematic equation:

Solving for the time is fairly straightforward. But what if I add in air resistance? What then? There is a problem. Air resistance is a force that depends on the velocity. This means that the force (and thus the acceleration) is not constant. That’s a big problem.

We can still solve this with a numerical calculations. In short, I can use a computer to model just a tiny time interval for a falling object. During this short time interval, the forces are roughly constant. Here is an older post that gives an introduction to numerical calculations. Also, don’t forget that my ebook (Just Enough Physics) has a whole chapter on numerical calculations.

Let’s just get to the calculation. Here is a model of a ping pong ball falling from a height of 10 meters. Actually, this is a Glowscript program so you can run it yourself and even edit it. Try it! In this calculation, I have a ping pong ball and a ball without air resistance dropped from the same height. In this plot, you can see that the ping pong ball hits after the no-air resistance ball with a time difference of 0.32

But this doesn’t answer the question: how high is too high? Of course, there isn’t just one answer to this question. The maximum height depends on how accurate you want your model. Here is the real plot that you want. This shows the falling time difference between an object with air resistance and one without for different starting heights. Actually, since larger starting heights will have larger times, I have plotted the fractional difference in times.

From this, it looks like a human drop height of about 160 meters would give a falling time about 10% different than that without air resistance. If you are just getting a rough estimate (like falling off a building), it would probably be fine to ignore air resistance. If you were dropping a ping pong ball instead, I would assume no air resistance for heights around just 4 meters.

But it’s not just about the falling time. Sometimes you care about the final velocity instead of the time. Could you just use the same cut-off heights for velocity that you do for time? I don’t think so. Let’s take a look at a falling human for example. If this human was falling off a building, near the end of the fall the air resistance would be much greater than it was at the beginning of the fall. However, this increase in speed at the very end might not make a huge difference in the falling time.

Here is a plot for the same objects showing the fractional difference in final velocities for with and without air resistance.

If I go with the same idea of getting just a 10 percent velocity error, the falling height for a human would 60 meters instead of 160 meters.

Homework

Ok, there are things left to look at – so I will give them as a homework assignment.

• What about the kinetic energy at the end of a fall? How high would a human have to fall so that there was just a 10 percent error in KE?


• Suppose I want to drop a ping pong ball in a lecture class to show that you need to include the air resistance in order to properly model its motion. How high should I drop this ball?


• I want to make some spheres made of wood (let’s say wood has a density of 900 kg/m³). Make a plot of drop height for a 10 percent fall time error vs. ball radius. As the ball gets larger, it’s mass to cross sectional area ratio changes. Bigger balls should be able to be dropped from a larger height with less error. How small should a wooded ball be such that if dropped from a height of 2 meters, the fall time is off by 10 percent?

That’s your homework.