When you drop a ball in the air, often the air resistance force is ignored. How high would you have to drop something so that the air resistance is significant?

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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.

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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.


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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).

See more: Light Cannot Escape The Intense Gravitational Pull Of A, Chapter 25 Review

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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: