When the drag force and force due to gravity are equal the terminal velocity is reached. Various factors are involved: 0.5dv^2AC=mg where d=air density, v=terminal velocity, A=surface area presented to the air, C=drag coefficient, assumed constant. The air density changes gradually with altitude, being densest near the ground. Although g does change with altitude, the change is very small compared to other factors so it can be regarded as constant at about 9.8m/s^2. A is constant as long as the falling object is not rotating. The mass, m, of the object is constant for a coherent object. From this v=√(2mg/dAC).
DERIVATION USING CALCULUS
We have to start by assuming the drag force at some velocity v is 0.5dv^2AC.
So if x is the altitude where x=0 is ground level, then gravity acts in the negative direction and drag in the positive direction. F=ma is Newton's Law and we can replace F with 0.5dv^2AC-mg, the net result of drag and gravitation. Initial velocity at time t=0 is zero so the only force is gravity -mg. As the object accelerates under gravity v increases. The acceleration a can be written dv/dt or v'.
So we have mv'=0.5dv^2AC-mg or v'=0.5dv^2AC/m-g=g(0.5dv^2AC/mg-1). Note that v is the vertical velocity upwards.
We can write this in the form of integrands, but before we do, let's simplify this a bit. We know which of these are constants so we can combine them into just one called k, where k^2=0.5dAC/mg. v'=g(k^2v^2-1).
∫(dv/(k^2v^2-1))=g∫dt.
Integrating by parts we have:
∫(dv/(2(kv-1)))-∫(dv/(2(kv+1)))=gt. (We'll insert the constant of integration later.)
ln(kv-1)/2k-ln(kv+1)/2k=gt; ln((kv-1)/(kv+1))=2kgt+C, where C is the constant of integration.
However, when v=0, t=0 so that would give us ln(-1)=C which has no meaning. So let's rephrase the original integrand: ∫(dv/(1-k^2v^2))=-g∫dt.
Now we have ∫(dv/(2(1-kv))+∫(dv/(2(1+kv))=-gt and
-ln(1-kv)+ln(1+kv)=-2kgt+C. This time when v=t=0, C=0, so ln((1+kv)/(1-kv))=-2kgt.
So (1+kv)/(1-kv)=e^(-2kgt).
After some time t=T the object reaches terminal velocity, but by definition the velocity stays constant and the object keeps falling at the same constant rate. If we imagine the object falls from a very great height so as never to reach the ground, t can be made infinite for all intents and purposes, and the right-hand side approaches zero. Therefore 1+kv=0 and v=-1/k=-√(2mg/dAC). The negative sign simply means the velocity is in the downward direction. A truer interpretation is that the falling object never reaches a terminal velocity no matter how far it falls, so the terminal velocity is in fact an asymptote.
v=-(1/k)(1-e^(-2kgt))/(1+e^(-2kgt)) is the actual equation for v.
For practical purposes, though, given sufficient altitude, over the range for which the quantities assumed to be constant are approximately constant, the terminal velocity has meaning.
