From: DENNIS_@ncsu.edu (DGA)
Newsgroups: rec.guns
Subject: Re: Shooting Straight Up
Date: 8 Jan 93 14:07:36 GMT
Organization: North Carolina State University

In article <1993Jan7.183906.24286@sol.ctr.columbia.edu> brown%ccit08@uunet.UU.NET (Jason Brown) writes:
#From: brown%ccit08@uunet.UU.NET (Jason Brown)
#Subject: Re: Shooting Straight Up
#Date: 7 Jan 93 23:59:54 GMT
#In article <1993Jan6.164626.14780@news.eng.convex.com> tford@convex.com (Tom Ford) writes:
## I heard a local news report on New Years Day about people shooting off guns in
## town to celebrate the new year, not a smart thing to do.  I was wondering
## what are the ballistics like for a falling bullet after it has been shot
## at near 90 degrees elevation.
## 
## Happy (safe) new year,
## Tom Ford
#
#Well,
#	If we step into the Physics room for a moment perhaps we can sort this
#one out.  Remember the simple fact that energy is conserved in a system. 
#At the moment the bullet leaves the chamber, we'll call that zero elevation,
#it has a certain amount of kinetic energy from the burning of the powder.
#As it climbs, this kinetic energy is converted to potential energy at such a
#rate that, at the apex of its flight (the top), Kinetic energy =0 and potential
#energy = kinetic energy at elevation zero.  So it stops, and now it starts to
#fall.  As it falls, it accelerates and the potential energy is turned back into
#kinetic energy.  Assuming that it went straight up and fell straight down, when
#it reached the chamber of the gun on the way back down, elevation zero, it
#would have the exact same amount of kinetic energy as when it left.  
#
#	What does this mean?  Well, if you shot the bullet, and then stepped
#into the spot you fired the gun from, when the bullet came down, you'd be dead.
#
#	Of course this little exercise does ignore a few things though.  Some 
#energy will be lost due to air friction and converted to heat, though not much
#if it is a very aerodynmic bullet.  Also, the shot has to go straight up and
#straight down as well.
#
#Hope this helps.
#
#--Jason Brown



Whoa there!

While my knowledge of physics is limited enough that I believe almost 
anything if presented with a good vocabulary, this one is a bit much.  I 
hope someone out there with the proper tools will explain this to Jason, 
me, or Jason and me.

The way I read the above, if I fire my hot little number straight up into 
the air with a muzzle vel. of say 3700 fps, and don't step to the side, then 
I am going to be beaned by a bullet traveling at very nearly 3700 fps a 
moment or two later.

Prior to this explanation, I was under the rather crude understanding that 
on the trip up, my projectile's battle with all of those nasty forces 
(gravity, friction, RAP music, and etc.) consumed the energy that was 
imparted to it by the bang.  On the way down all I had to worry about was 
the force imparted to the projectile by gravity (minus friction, RAP music, 
and etc.).  Instead of being killed by a projectile traveling at very nearly 
3700 fps, my demise would occure as the result of being struck by a 
projectile traveling at a mere 300 or so fps.  This of course would be much 
neater.

Someone set me straight.  Consider my formal training is as an economist.  I 
am prepared to assume anything.

Thanks

DGA