If an AR15 223rd, 55gr FMJ,was fired into the air at what velocity, after expending its foot pound energy,would it fall to the ground & what danger does it pose to life and property? I know that angle fired makes a difference.

Like you say:  "expending"
Is a kinda a matter of friction.

well, if fired straight up, it would probably land with a force of about 250 fps, about the same as daisy red ryder bb gun PROBABLY.  barely enough to break skin, that's assuming a straight up and down trajectory with 0 wind deflection, but realistically, it's like a richochet, you can never tell, there's always an exception to the rule.
you have to remember that on the way down, it will only achieve terminal velocity, (roughly 140 mph).

Quoted: If fired straight up I think it would strike the ground at about the velocity it left the muzzle. Of course I only got an 85 on my high school physics final.

Yeah, if the muzzle were at ground level, right?
(I got >85)

too many variables..  Fired straight up, it would stop and gain speed while falling until it reached terminal velocity..  Can't remember what that is...

Well if you read Naked-gunmans post about firing his AR into the air to scare off intruders alot of people were very concerned about the rds returning over a populated area. What is the returning rds potenial for damage?

Isn't the terminal velocity of any object unique to that object; and dependent on a whole slew of variables like its shape, wind resistance, weight/density, the atmospheric conditions, etc.  This would be even harder to calculate for a bullet because the projectile would probably be tumbling as it fell, so its ballistic co-efficient and shape/resistance of the bullet would constantly be changing in relation to what part of the bullet was currently facing straight down.  You would have to average all the resistances of the various profiles that it could present, considering that any given projectile could tumble differently than any other particular one.  It would be hard to get more than a close approximation.

I saw a calculation for a penny dropped from the Empire State Building and the approximate terminal velocity was low enough that it would not be as dangerous as normally believed.

One way I have read about to determine terminal velocity (approximately) for a complex object is to attach it to a string, hold the string outside a vehicle with the object hanging straight down on a calm day.  Have the driver begin to drive, when the object is hanging at a 45 degree angle the effects of wind are about equal to those of gravity and the speed displayed on the speedometer is somewhat close to terminal velocity for that object.

Of course once again, with complex shapes it will be difficult to determine the effects of the various possible resistances if the object were actually falling and not held in one contant position by the string.

For example a screw driver dropped carefully and exactly vertically on a calm day would probably achieve a higher speed than the same screwdriver dropped horizontally or haphazardly tossed.

At least thats what I think.


I

I dont know about .223 but a .45 230gr FMJ will punch through the roof of a car.

I know, I was standing 5 feet away when it came in.

PS It was a 65 Falcon. They used real steel in those days.

Quoted: If fired straight up I think it would strike the ground at about the velocity it left the muzzle. Of course I only got an 85 on my high school physics final.

HA-HA!
85.....

Quoted: ...barely enough to break skin...

HA HA HA HA HA...I hope that was a joke because it surely is funny.  

I will ditto what tc6969 said, as many years ago my neighbor had a 9mm ball come crashing through her carport (corrugated steel, mind you) and penetrate her ~1982 Lincoln Continental's hood.  That bullet was sitting in the hood after punching through that carport straight down.

Unfortunately, the neighbor I grew up in had a lot of illegal hispanics who loved to shoot their pistols in the air on holidays.  This was the result.

Also, a good friend of mine who knows little to nothing about firearms said that in HIS neighborhood the same thing would happen...but one 4th of July they saw sparking on a large concrete slab (it was a set of outdoor basketball courts at the local Jr. High) due to bullets striking.

Barely break the skin...Hooo, thanks for comic relief!

Right!!  It's a matter of friction..  A steel ball and a feather, dropped in a vacuum would fall at exactly the same speed...

Quoted: If fired straight up I think it would strike the ground at about the velocity it left the muzzle. Of course I only got an 85 on my high school physics final.

In a vacuum that would be correct.

As others have said, it depends on the terminal velocity of the bullet.  Wasn't there a kid killed by a falling bullet after some New Year's of 4th of July shooting last year?

Quoted: ...but one 4th of July they saw sparking on a large concrete slab (it was a set of outdoor basketball courts at the local Jr. High) due to bullets striking.

Brass, copper, or lead causing sparks?

Quoted: Quoted: ...but one 4th of July they saw sparking on a large concrete slab (it was a set of outdoor basketball courts at the local Jr. High) due to bullets striking. Brass, copper, or lead causing sparks?

Wolf Ammo

Quoted: Quoted: Quoted: ...but one 4th of July they saw sparking on a large concrete slab (it was a set of outdoor basketball courts at the local Jr. High) due to bullets striking. Brass, copper, or lead causing sparks? Wolf Ammo

RIGHT ON!!  We shoot a gravel wash with wolf at dusk..  It's kinda neat...

You enter your own figures.

www.processassociates.com/process/separate/termvel.htm

Or, try this one.
www.explorescience.com/activities/Activity_page.cfm?ActivityID=27

Gentlemen..gentlemen..its simple
the only thing affecting the bullet back down...is gravity..(lets neglect air friction..not like we care THAT much do we?)

Basic newtonian mechanics..F=MA

Just what you need:

www.loadammo.com/Topics/March01.htm
www.loadammo.com/Topics/May01.htm

terminal velocity will be ~300 fps (i.e. much more than a pat on the back, but not quite equal to a handgun shot)

maximum range will be somewhere between 1 and 2 miles, if fired at 35 degrees angle (max range occurs at less than 45 degrees due to air resistance)...

Quoted: Gentlemen..gentlemen..its simple the only thing affecting the bullet back down...is gravity..(lets neglect air friction..not like we care THAT much do we?) Basic newtonian mechanics..F=MA

Uh, I think that this guy got a 50% in the class.

If I recall correctly the Army says a projectile at 300 fps is the minimum required to create a casualty.

For comparison, the slingshot I used as a kid would only do about 250 - 275 fps, but it would take the head off a mourning dove.  Now humans are build a little more ruggedly that a mourning dove, but I, for one, will not volunteer to get hit by a falling bullet.

Kent

Quoted: If fired straight up I think it would strike the ground at about the velocity it left the muzzle. Of course I only got an 85 on my high school physics final.

Only on the moon or another body without an atmosphere.

My compound bow sends my arrows out at 230 fps.
Broadhead or field point, or even a blunt tip, I would not want to get hit with an object traveling at that speed.

Mass has a lot to do with the damage created. If a plastic bb hits you at 200 fps, the energy will be released very rapidly, due to it's low amount of mass. It's energy would be absorbed by the skin through the eleastic property of the skin. A lead bb of the same proportions would need a longer time to release it's energy...meaning it wouldn't just bounce off the skin, it would more than likely penetrate before it looses it's enegy.

True both objects would be traveling at roughly the same speed (to marginal to be a factor), but the lead bb has more kinetic energy due to it's higher density of material.

Quoted: If fired straight up I think it would strike the ground at about the velocity it left the muzzle. Of course I only got an 85 on my high school physics final.

Nope

The muzzle velocity would be much greater than terminal velocity.  As it goes up, it slows down until it stops, then gravity pulls it back down.  Gravity ain't as strong as the exploding gasses in the case, so the bullet has no push when it comes back down.  It only has it's weight as a potential, and is only resisted by wind.

Just another dumb question, if I fired straight up, how long will it take to get back to the ground leaving the muzzel at 3100fps?  And how high would that bullet go?

Sixty pound feet of energy is what was determined to be needed to produce a disabling wound.. The US M2 Ball round fired vertically will fall with an average velocity of 300 feet per second, and an energy of 30 pound feet..(And usually be returning sideways, or base first)"Thus, service bullets returning from extreme heights cannot be considered lethal, by this standard "(60 ft/lbs to induce a disabling injury)

As to penetration, I once repaired the swamp coolers on a business owned by some friends.. The roof was covered with spent bullets, ranging from copper clad shot, to .45 caliber projectiles..The worst damage was minor denting to the tops of the cooler housings..Some had sunk into the roof's tar, but considering many had probably been fired on the Fourth, the tar would have been nice and soft..None had penetrated, either the cooler's thin metal, nor the roof..

I believe that many of the "incoming round" incidents are the result of idiotic/drunken gun handling, and negligent discharges..Not bullets whistling in from the blue.

Meplat-

Ask the Palestinian. They seem to fire in the air all the time. I always wonder where the  bullets go.

Quoted: I believe that many of the "incoming round" incidents are the result of idiotic/drunken gun handling, and negligent discharges..Not bullets whistling in from the blue. Meplat-

I said this before, but was jeered at and called an irresponsible idiot for purposely trying to rain down death and destruction on the children of the world.

A 55 grainer would do little damage at worst, and if you live out in the country, your chances of hitting ANYTHING is infitesimally small.  Even hitting your own house would be tricky if you tried.

Think about it.  You would probably be able to fire a round 5000 feet high perhaps?  In order for it to land on your roof, it has to return another 5000 feet and hit the target.  In other words, you would have to hit your target two miles distant, and without aiming.  How easy is it to determine exact straight up into a blue sky?  A couple degrees off, and you will miss.  The wind makes this even harder.

I wouldn't recommend shooting into the air unless you have a really good reason to.  The more you do it, the better your chances of causing an accident are.

Wellll, I've never had a 230gr projectile fall on me from the sky, but I have had steel shot "rain" down on me while duck hunting in a swamp a few times. That is just what it sounds and feels like, a quick burst of ~hard rain. One of the few times I was happy to be wearing glasses. And a boonie hat.

More of a concern to me is dodging pheasants or geese. I've had to step out of the way a time or two as I've introduced a fast flying oncoming bird to a load of birdshot. Makes for an easy retrieve, however.

Quoted: Just another dumb question, if I fired straight up, how long will it take to get back to the ground leaving the muzzel at 3100fps?  And how high would that bullet go?

Not a dumb question, but a classic example of a one dimensional kinematics problem:
For the sake of simplicity, the muzzle is at ground level:

-initial velocity (vo)=3100ft/sec=944.88m/s (I like SI units)
-acceleration(a) due to gravity=9.8m/s/s
-So, to derive the formula that we'll use:
--a=final velocity(vf)-vo/time
--So, at=vf-vo
--So, vf=vo+at

Because the definition of average velocity is:  average velocity(va)=(final position of bullet(xf)-initial position of bullet(xo))/time, we can rewrite this equation as:
xf=xo-va*t

velocity changes at a uniform rate, so:
va=(vo+vf)/2

We can now combine these last two equations with our first derived equation to find that:

xf=xo+va*t=xo+((vo=vf)/2)*t=xo+((vo+vo+at)/t)
OR:
xf=xo+vot+(a*t*t)/2

Plugging in the numbers:
the position of the bullet, in our simple example, is the same at the beginning and at the end, so:

xf=0
xo=0
vo=944.88m/s
(t unknown)
a=9.8m/s/s

0=(944.88m/s)*t+((-9.8m/s/s)*t*t)/2
factor out a t:
((944.88m/s-4.9m/s/s)*t)*t=0
t=944.88m/s/4.9m/s/s=192s

Maybe someone could check that for me, and correct all the typos?

10 or 12 years ago,some guys were shooting in the woods a mile or two from Carowinds(theme park).
One of them fired in the air or over their backstop(been a long time).

Anyway, a bullet(7.62x39) hit a girl in the wave pool. Killed her.

I wouldn't want to get hit by a "stray" round, coming down.

Caveman- That's NOT the same.. The question regarded high trajectory returning bullets.. Your example is in the realm of low trajectory, extreme range incidents.. The fact that it cleared the berm indicates a what, 30 degree angle of departure? Big diffrence from a 90 degree angle of departure, especially when one is considering the velocity, and energy the bullet retains..

Meplat-