# Falling and Air Resistance

## Discussion:

When an object falls in air, the air exerts an upward air resistance force (sometimes called a "drag force" ) on it. As the speed of the falling object increases, so does the air resistance force. When the air resistance force equals the weight of the object, it stops accelerating and falls with a constant terminal velocity. Light objects - feathers, pieces of paper, and coffee filters exhibit this behavior quite readily - since they have a large surface area and a small weight, they reach their terminal velocity very quickly.

Certainly, the air resistance force on an object depends on its speed - but how? Suppose that the air resistance force is proportional to the objects velocity, then, at terminal velocity:

Fair = k v = W

where k is a constant that depends on the shape of the object and the density of the air, or:

v = W/k

since the distance that an object moves (at constant velocity) equals velocity times time,

This equation says "distance fallen is proportional to weight" - in other words, an object twice as heavy will fall twice as far in the same time, or an object three times as heavy will fall three times as far in the same time, etc.

On the other hand, what if the air resistance force is proportional to the square of the speed? In this case, at terminal velocity:

Fair = kv2 = W

(where k is a constant) or:

So if air resistance is proportional to v2, then distance fallen is proportional to the square root of the object's weight, so an object twice as heavy would fall about 1.4 times as far in the same time, and object three times as heavy would fall about 1.7 times as far in the same time, etc.

So, is the air resistance force on a typical falling object proportional to its velocity, or the square of the velocity?

## coffee filters meter sticks stopwatch (?) Procedure:

First of all, notice that two coffee filters placed together have twice the weight as one coffee filter, but about the same surface area. Three coffee filters have three times the weight as one coffee filter, etc. This means that the appropriate unit of weight in this lab would be "coffee filters", as in "1 coffee filter", "2 coffee filters", etc.

From here, there are a couple of reasonable ways to proceed. The "straightforward" way would be to drop coffee filters a known distance and time their fall. A graphical analysis (starting with weight vs. time of fall, perhaps) of carefully collected and recorded data might reveal whether the drag force is proportional to velocity or velocity squared.

Another approach would be to drop a single coffee filter a known distance (a meter, say) each time, and determine the height from which a double (or triple, whatever) coffee filter must be dropped to hit the floor in the same time. (If Fair is proportional to v, this would be 2 meters...) You wouldn't have to mess with a stopwatch this way, for one thing. Here, you would be analyzing weight vs. distance fallen.

## Analysis:

Which model for air resistance (if any) does your experiment support? Why do you think so?

last update October 30, 2002 by Jerry Stanbrough