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Raindrops

Part 2: Falling Bodies with Air Resistance

We now investigate a more sophisticated model for a falling body, one that takes into account the resisting force of the air through which the object falls. The usual physical assumption is that the force of air resistance is proportional to some power of the velocity, but the particular power (first, second, other) depends on the particular object.

We consider raindrops falling from a cloud 3000 feet above the ground. If the raindrop is small, say a drop of diameter 0.00025 feet (or 0.003 inches, a size found in a drizzle), the force of air resistance is modeled well by a multiple of the first power of the velocity. In other words, the resisting force can be described by

Fr = -k v

for some constant k. (The minus sign indicates that the force is in the direction opposite to the velocity, i.e., upward.) When combined with the force of gravity,

Fg = m g,

this yields the total force on the raindrop:

F = Fg + Fr.

We recall Newton's Second Law of Motion:

F = m dv/dt.

Equating our two formulas for the force F and dividing by m, we find a new differential equation for velocity:

dv/dt = g - (k/m) v.

We'll let c represent the quotient k/m. When we attach our initial condition, v(0)=0, we obtain our new initial value problem:

dv/dt = g - cv, v(0)=0.

Experimental evidence gives an approximate value of 52.6 sec-1 for c, when distances are in feet and the drops are drizzle size.

Guessing a formula for v as a function of time may be more difficult for this problem than actual calculation was in Part 1. You may or may not know a systematic way to solve this type of problem at this point. In the next part we will take a different approach and use a numerical technique called Euler's Method. This technique will only approximate the desired solution, but it has the distinct advantage of applicability to any initial value problem.

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Last modified: October 14, 1997