DA> JD>The formula for calculation the velocity of any object falling in a
DA> JD>on this planet is: 32 feet per second squared times the time that th
DA> 
DA> JS>Time in seconds squared * 32.16fps (in Chicago, latitude 42ø) actual
DA> 
DA> JS>JD>object has fallen.
DA> 
DA> You both flunk.  :-(  Acceleration is delta-velocity divided by
DA> delta-time.  Delta = change in.  So velocity, when starting from
DA> rest, is merely acceleration times TIME.  There's nothing squared
DA> in that. (Assuming a vacuum and constant acceleration, etc.)
While you are correct about the definition of velocity, Jackson Dryden
was correct too.  Perhaps you misunderstood him.
I understood him to say that v = a * t, where v is velocity, a is
acceleration, and t is time.  I believe that's also what you said.
    a = 32 ft/sec^2
Maybe his use of the units "feet per second squared" confused you.  His
use of the term "squared" related to the units and not the equation.
Actually at sea level, the gravitational constant is 32.1740485564...
ft/sec^2.
If the object was at rest and released in a vacuum at sea level, in one
second the object would be traveling at a velocity of 32 fps.  In two
seconds it would be at 64 fps, at three seconds, 96 fps, and so on.
(OK, so I rounded.)
DA> Perhaps you guys were thinking DISTANCE.  DISTANCE = .5 times
DA> ACCELERATION times TIME^2.
That's correct.  The only time you were wrong was when you said Jackson
was wrong.  :)
DA> I don't have the formulas for calculating terminal velocity in
DA> real life. Maybe Bill Frenchu will share them with us. Given the
DA> bullet's sectional density and ballistic coefficient, it should
DA> be able to be determined.
I don't know what the bullet's "density" would have to do with it, but
the mass, the drag coefficient, the air density, and the cross sectional
area would have a bearing on the terminal velocity.  In addition, the
drag coefficient and cross-sectional area would change if the bullet was
tumbling rather than falling in a fixed relative position.  It sounds
like a good fluid dynamics problem.  :)
The terminal velocity should be reached when the drag force equals the
gravitational force.  In terms of an equation, it is as follows:
0.5 * d * C * A * v^2 = m * g
where d is the density of air, C is the drag coefficient, A is the
cross-sectional area, v is velocity, m is mass, and g is the
gravitational constant (32 ft/sec^2).  The left side of the equation is
the drag force, while the right side is the gravitational force.
Solving for the velocity yields the following equation:
v = sqrt( (2 * m * g / d * C * A) )
Of course you must use the correct units in the equation, and the drag
coefficient C must be found experimentally.
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