"SimonShabtai Evan" wrote in message
...
No, I'm not mistaken about a "little slower". This is a relative term
with wide variation in value; in other words, I have not done the
calculations but the velocity will be slower.
If you real want the velocity of the bullet apply the following:

(1) v = v(0)+at
(2) x = x(0)+1/2(v(0)+v)t
(3) x = x(0)+v(0)t+ 1/2at^^2
(4) v^^2 = v(0)^^2 + 2a(x-x(0))

Also note that these calcuations (from Physics 101; Halliday and
Resnick/Wiley and Sons Inc) do not include atmospheric losses.

have fun.
S. Evan

Tumbling which way? Bullets tumble when they lose a percentage of their rpm.
Bullets shot straight up always tumble. There go the old aerodynamics...

FWIW, before WWII the US Army tested .30 cal military bullets shot straight
up. They returned to Earth at a velocity of 200 - 220 mph. It didn't matter
how high they were shot, what their initial muzzle velocity was, etc.

Specific sectional density doesn't matter much. Aerodymanics go out the
window once they start to tumble. What matters is their mass relative to
*effective* projected cross-sectional area, times some factor also related
to mass. This is a backhanded way of getting at the Reynolds number for a
given-size bullet. The "effective" area is a factor of shape and the way it
tumbles.

I understand that the terminal velocity of a spread-eagled human is
something like 120 mph. It doesn't matter if you fall from 2,000 feet or
50,000 feet, you hit the ground at around 120 per. Does anyone have better
data on that?

Ed Huntress