Re: The exact formula for terminal velocity with its algerbraic symbols

Richard,

I think I know what you are looking for, but in physics jargon it's really 
an approximate equation for terminal speed (I say speed because velocity 
describes both speed and direction).

The two dominant forces in free-falling through the atmosphere are drag and 
gravity.  (Atmospheric) Drag is due to what we call (skin or surface) 
friction.  Long story short, the typical formula for atmospheric drag is

D = (1/2)*C*A*p*(V^2)

D is the drag force, in Newtons (or pounds)
C is the coefficient of drag, or drag coefficient, which is unitless
     C is often written as Cd where "d" is a subscript
A is the area of the body's surface over which the air is flowing
p is the density of the air, in kg/(m^3) or slugs per cubic foot
V is the speed of the air flow relative to the body which is experiencing 
the drag
     V is expressed in meters per second (m/s) or feet per second

The units for D is then ( kg/(m^3) * (m^2) * (m/s)^2 ) = kg*m/(s^2) = 
Newtons
or ( slugs/(ft^3) * (ft^2) * (ft/s)^2 ) = slugs*ft/(s^2) = pounds

note: a slug is the imperial unit for mass, just like the kilogram (kg) is 
the SI unit for mass

note: the density of air is not constant, it changes over time and 
location.  You can read more about this in the references, below.

The force exerted by gravity is called weight.  The formula is

W = m*g

W is the weight, in Newtons (or pounds)
m is the mass of the object, in kg (or slugs)
g is the acceleration due to gravity, in m/(s^2) (or ft/(s^2))

note: I'm using the carat "^" for exponentiation.  a^b is "a" raised to the 
power of "b", so that 2^2 is 4, 4^2 is 16

note: acceleration due to gravity changes with location.  This is often 
ignored for problems like what you're describing, but if you double your 
distance from the center of Earth, then the acceleration due to gravity is 
one-fourth what you felt before.

If there were no atmosphere to slow you down, the force of gravity would 
continually accelerate you until you collide with something solid (like the 
earth).  However, the atmosphere does exists and it retards your motion 
with the force of drag.  Therefore, the resultant force you feel towards 
the earth is weight minus drag (W-D) and when your speed increases to the 
point that the drag you feel equals your weight, you will accelerate no 
more.  This then, is your terminal speed.  And the equation to find it is:

W = D
m*g =  (1/2)*C*A*p*(V^2)

and we can rearrange this with some algebraic manipulation to get

V^2 = 2*m*g/(C*A*p)
or
V = ( 2*m*g/(C*A*p) )^(1/2)

for more information you should look to:

Anderson, John D., Introduction to Flight, New York: McGraw-Hill, 1989.
"standard atmosphere" http://www.britannica.com/bcom/eb/article/0/
0,5716,71190+1+69392,00.html
"terminal velocity" http://www.lerc.nasa.gov/Other_Groups/K-12/airplane/
termv.html
"atmospheric pressure" http://www.britannica.com/bcom/eb/article/
8/0,5716,10248+1+10121,00.html
"PROPERTIES OF THE U.S. STANDARD ATMOSPHERE 1976" http://www.pdas.com/atmos.htm
"WebWeather Standard Atmosphere Computations Calculator" http://www.webweather.com/
toolbox/atmosphercalc.htm
"Standard Atmosphere Computations" http://aero.stanford.edu/StdAtm.html
  http://nssdc.gsfc.nasa.gov/space/model/atmos/us_standard.html
"The drag coefficient" http://www.lerc.nasa.gov/Other_Groups/K-12/airplane/
dragco.html
"terminal velocity in weather" http://www.nssl.noaa.gov/~cortinas/1014/l16_1.html