Terminal Velocity!

One thing I always wanted to learn how to do since I first started getting into physics was calculating the terminal velocity of an object falling. It turns out it really isn't all that difficult!  It requires you to solve a second order non-homogeneous differential equation. It sounds like a mouthful but it turns out that it isn't that bad!

First thing we must realize is that the "drag force" on a falling object is directly proportional to its velocity. This makes sense because the faster the object is traveling, the more air it has to push through, and hence the more the air slows the object down.  Therefore..

The proportionality constant is itself proportional to the cross sectional area of the object in question, and the density of the air.  Again this makes sense because the denser the air is, the more resistance the object feels when falling,  and the cross sectional area in question is the area in which the air is "running into" the object.  So if it was a cube falling, the bottom face would be the area in question.  We will leave this as "c" until the end to simplify things a little bit.  We now just need to use newtons second law.  Remembering that acceleration is the derivative of velocity, we can obtain..

Let's run through what just happened.  We first summed the forces in the vertical direction (because horizontal contributes nothing to the velocity in question.), then we expressed  the variables in terms of velocity.  After doing this, we obtained a second order, non-homogeneous differential equation.  From here we used an integrating factor to express the left hand side as the product rule of differentiation and integrated both sides.  We now have to solve for the unknown constant, therefore we need to come up with an initial condition.  The initial condition is fairly trivial in this case, the velocity at time zero, is zero! So v(0)=0.  Plugging this into the differential equation and solving for c, and then remembering what we said earlier about c being proportional to area and density we obtain..

We now test this model with intuition.  But before we do this, we need to examine the limiting case, namely the value of the terminal velocity.

If we have a very large mass, we would expect the terminal velocity to be obtained longer than a light mass, we would also expect it to be a greater value.  

The mass is squared in the numerator of the limiting term, therefore an object of the same cross sectional area but the same mass falling through the same air density will fall four times faster.  This also happens at a later time because the mass is in the denominator of the exponent term in the exponential.  Therefore it takes longer for that exponential to equal zero.

Our next intuition is that an object with a greater cross sectional area would have a lower terminal velocity and achieve it much faster than an object with the same mass but less area.  Our model predicts this exactly, the area is in the denominator of the limiting term, meaning, if it is very large, the terminal velocity will be very small.  It is also in the numerator of the exponential term, meaning that when area is very large, the whole exponential term tends to zero much faster.

Our last intuition is that the greater the air density, the slower the terminal velocity will be and the faster it will be reached since there is more air to slow the object down.  (Note! This object can also be used to model an object falling into a fluid, although the model of F=cv² works much better for this situation!)  Our model predicts this behavior exactly.  Then the density is very large, the terminal velocity is very small.  It also occurs much faster because the density is in the numerator of the term in the exponent.

So there you have it! What you can do now, is estimate your cross sectional area in various falling positions and use 1.00 as the density of air.  You would expect that falling in a nose dive would make you fall much faster than falling with your arms and legs spread out wide, which this model predicts exactly!