"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.

FWIW Frictional losses are very significant at high velocities(i.g. shuttle
reintry). When I was in infantry school at Ft. Jackson during the last
'action' in S.E. Asia I had an opportunity to get one of those info cards
that came in each case of 50 cal. machine gun ammo. On that card, range,
angle, velocity data were given WITH atmospheric effects. If I just used
the equations listed above and optimum 45 deg angle I calculated the range
of the 50 cal. to be IIRC miles but the tables showed about 20% what I
calculated w/o allowances for air resistance. If one is really interested
in this stuff you might look at some of the work Galileo(1564-1642) did with
cannon trajectories years before Newton(1642-1727) Just Google _Galileo
projectile motion_ or some such.
Larry