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Old 12-04-08 | 03:31 PM
  #52  
nafun
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Joined: Dec 2007
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From: Austin
Originally Posted by <3 2 Ride
For today's lesson in physics:

"A free falling object achieves its terminal velocity when the downward force of gravity (Fg)equals the upward force of drag (Fd). This causes the net force on the object to be zero, resulting in an acceleration of zero. Mathematically an object asymptotically approaches and can never reach its terminal velocity.

As the object accelerates (usually downwards due to gravity), the drag force acting on the object increases. At a particular speed, the drag force produced will equal the object's weight (mg). Eventually, it plummets at a constant speed called terminal velocity (also called settling velocity). Terminal velocity varies directly with the ratio of drag to weight. More drag means a lower terminal velocity, while increased weight means a higher terminal velocity.

Mathematically, terminal velocity is given by

Vt = sq rt (2mg/ρACd)

where

Vt = terminal velocity,
m = mass of the falling object,
g = gravitational acceleration,
Cd = drag coefficient,
ρ = density of the fluid through which the object is falling, and
A = projected area of the object. "

I hope that answers your question for you.
Just to preempt the people who will inevitably claim this only matters at ridiculous speeds, like over 100mph, we can factor in the slope of the hill:

Vt = sqrt((2*m*g*cos(θ))/(ρ*A*Cd))

where θ = angle of decline

Last edited by nafun; 12-04-08 at 04:06 PM.
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