I am not sure what part of what I said is bothering Jack and Brian.
(Probably different parts.)
Yes, if the bullet is launched with some horizontal component, then the
velocity is a vector with both vertical and horizontal components. Yes,
the air friction direction is opposite the actual velocity vector; air
friction is not just vertical.
There is an additional force in the z-direction (gravity) that is not
present in the x or y-direction. Therefore it is possible that the
z-motion can reach a point that the z-component of the acceleration is
zero. I'm calling that the terminal velocity. As Jack says (without
using x,y,z language), the z-component of the air friction matches
gravity, and there is no net z-force, hence no z-acceleration. The
z-velocity at that point is what I would call terminal.
In the x or y-direction the only force is air-friction, and that will
continue to create an x and y-acceleration until the x and y velocities
are zero. When I say there is "some horizontal velocity in addition to
the terminal vertical velocity" I am assuming the z-acceleration has
become zero or near zero (terminal velocity, or terminal z-velocity if
you prefer) but I am not necessarily assuming the x and y-velocities
have accelerated to zero. Maybe they have, maybe they haven't.
Yes, I assume that because air friction is opposite the actual velocity
vector that it is probable the z-velocity at which the z-component of
the drag equals the weight depends some on the x and y-velocities.
I still maintain that when the bullet is fired within a few degrees of
vertical that the z-velocity of landing is near the z-velocity of a
simple free fall from whatever height the bullet reached, and I assume
height reached is enough that the z-acceleration just before landing is
darn-near zero.
An M16 (not what most people have) has a muzzle velocity of about 1000
m/s (about 2000 mph). If shot at an angle of 80-degrees or more, the
horizontal launch velocity will be less than 350 mph. Can we do the
following *rough* calculation... If the terminal velocity of a dropped
M16 bullet is 100 mph, then the horizontal drag of a bullet traveling
with a horizontal component of velocity in excess of 100 mph will be
sufficient to cause horizontal acceleration of at least -g. If so, it
will take less than 5 seconds for the horizontal component to accelerate
from 350 mph to 100 mph. (I acknowledge this might require tumbling
which is not going to be happening right from the start.) The bullet is
going to be in the air for several tens of seconds. It seems to me that
the horizontal component of the velocity upon landing is going to be
pretty small if the bullet was fired "mostly up." Maybe a few mph. If
the landing velocity is roughly 100 mph down and 10 mph horizontal, then
the impact velocity is about 100.5 mph. Even if the horizontal velocity
is 30 mph (and the vertical velocity is 100 mph) the impact velocity is
only 104 mph.
It seems to me that bullets are only really dangerous when fired at
small launch angles. When this is the case they reach their target very
quickly and still have high velocity. As the launch angle increases,
the time-of-flight increases dramatically, acceleration from drag has a
lot longer to act, the velocity at impact is a lot less.
Michael D. Edmiston, Ph.D.
Professor of Chemistry and Physics
Bluffton University
Bluffton, OH 45817
(419)-358-3270
edmiston@bluffton.edu