# Terminal Velocity Equation

1. Aug 28, 2004

### Jamez

i'm looking for an equation to calculate terminal velocity. Does any one know it? and can u please post in here please.

2. Aug 28, 2004

### arildno

That would depend upon how you choose to model the air/fluid resistance.

3. Aug 28, 2004

### HallsofIvy

Staff Emeritus
One common model is that the resistance force is proportional to the speed.
Under that model, an object falling, under gravity has acceleration -g+ kv (k is the proportionality constant, v the speed. Since that is a function of v, it give the linear differential equation mdv/dt= -mg+ kv. The general solution to that is v(t)= Ce-kt/m-mg/k. For very large t, that exponential (with negative exponent) goes to 0 and the "terminal velocity" is -mg/k.

Another common model is to set the resistance force proportional to the square of the speed. That means the net force is -g+ kv2 and v satisfies the differential equation mdv/dt= -g+ kv2. That's a non-linear differential equation but is separable and first order. We can integrate it by writing
Integrating both sides, we get (1/2√(kg))ln((√(k)v-√(g))/(√(k)v+√(g))= mt+ C. For large t, the denominator on the left must go to 0: the terminal velocity is -√(g/k) which, you will notice, is independent of m. This model is typically used for very light objects falling through air or objects falling through water.

4. Aug 28, 2004

### da_willem

A falling object on earth is subjected to a downward force $F_g=mg$, while it's air resitance constitutes an upward force often modelled by

$$F_w=\frac{1}{2}C_D \rho A_F v^2$$.

With $C_D$ a constant (drag coefficient) that models how aerodynamic the object is, for most object of the order 1 (for a raindrop for example ~0,5), $\rho$ the air density, $A_F$ the frontal area of the object (perpendicular to the direction of motion), and $v$ the velocity of the object.

At terminal velocity the force on the object is zero (otherwise the object would accellerate!) so you can equate both force-equations yielding:

$$mg=\frac{1}{2}C_D \rho A_F v^2$$
$$v=\sqrt{\frac{2mg}{C_D \rho A_F}}$$

5. Jun 23, 2006

### hoppingbuffalo

Error in post 3

Indeed, terminal velocity depends on mass. You forgot the FORCE due to gravity is mg, not g.