How long does it take a bullet to come down, if you shoot straight up?

I've always wondered if an F-16 in a full-throttle vertical dive, fired it's 20mm Vulcan cannon, would it shoot itself down when it eventually caught up with the cloud of bullets?

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Yes. There have been several incidents where a fighter jet caught up with cannon fire during a dive, and subsequently damaged the jet. Generally speaking it's not as damaging as if being shot at directly, since the relative velocities of the projectiles are much closer than if the cannon were purposely fired at the jet.

Stand Skinny.................

I remember watching that epsiode of mythbusters and they said that as long as the bullet was fired perfectly straight up then when it came back down it wouldnt have enough energy to kill someone, however they stated that if it was fired at more than a few degrees past 90 degrees then it could remain lethal when it came back down. If I remember right they fired a Garand and 9mm handgun and both rounds barely penetrated more than 2" into dirt. My memory of it is a bit sketchy so hopefully someone else remembers it aswell.

Chris

thats the way i remember the show also.

2 rules about wive's:
#1--they make all the rules
#2--they can change the rules at anytime and they do not have to tell us.

I'd have to dig the book out, but in Hatcher's Notebook, the General talks about experments he did between wars. His team fired 30/06 rifle bullets straight up from a platform in the middle of a fair sized lake and timed the bullet's flight. Most of the time, the bullets missed the lake completly. Once, a bullet hit the roof of the shelter they were standing under and bounced into the water. Once a bullet hit the side of a rowboat tied up to the platform. It made a small dent. From the shape of the dent, he concluded that the bullet hit base first.

When I lived in Tuscon, some friends and I tried the same thing with .45 pistols. Firing as close to straight up as we could, we never saw or heard a bullet fall. Probably fired about 30 rounds. Nada.

Hummm...And your intellectual level? Wonder where they came down?

I remember watching that epsiode of mythbusters and they said that as long as the bullet was fired perfectly straight up then when it came back down it wouldnt have enough energy to kill someone, ...

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As a complete non-physicist, I that they concluded that the bullets hit "terminal velocity" on the way back down, and that terminal velocity was substantially less than muzzle velocity.

Mike

No problem, Bushy, we were over 20 miles outside of town. Flat desert. We'd sit in the car and look for puffs of dust. Never saw a thing.

If the bullet is shot straight up an a 90* angle, it will tumble on the way down and not be a problem.

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Mythbusters disproved this. It doesn't tumble
The time it takes to go up is dependant on the velocity. That decreases by 32'/sec/sec. That will determine the maximum altitude, minus a bit for air resistance. On the way down, it will max out at about 120 mph. and assume a sideways orientation.
When shot straight up, the bullet did not penetrate anything with enough force to kill someone. (2" deep in the hard desert sand).
As we heard, the one NOT shot straight up may kill someone.

O K...Just checkin'...

If you're given the mass of the bullet and it's initial velocity, this isn't all that hard of a problem if you ignore air resistance and only care about gravity. Someone post some fictional or real numbers and I'll do it, if you want.

i would say longer than that. give it about a minute. mile long shots can take sometimes as long as 10 seconds (thats a guesstimate, dont shoot me). thats parallel with the ground. one would think that straight up would take long to hit.

At terminal velocity, how hard does a 9mm 125grn bullet hit? I have no idea, but I have heard of people being hit in the head and the bullet breaking the skin. I have not heard of many people being killed by a bullet traveling at terminal velocity, however I bet it is possible. The best bet is to stay in the house. I would expect the bullet to reach the surface of the earth within 2 minutes. It does not matter what powder charge the shell has because we are assuming the bullet is shot straight up and reaches terminal velocity.

No one mentioned the treadmill

anyone remember calculus

1150 feet per sec will reach -0- speed at 8 seconds. takinig 8 seconds to fall back tand stike the ground at a a speed of 246 feet per second [167 mph]. air resistance not factored in. terminal velocity would likely be closer to 125 mph.

16 seconds 'flight time'

who can figure the height--the apex ??

air resistance not factored in

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Now I am a little confused - I thought that without air resistance, a vertically fired bullet will reach the ground at exactly the speed with which it left the ground.

Mike

If you assume no air resistance,
v = u + at
s = ut + 1/2at^2

v is velocity, u is starting velocity, t is time, a is acceleration, s is distance.

So take your muzzle velocity u, make v = 0, remember that a is negative (bullet slowing down to v = 0) and get t=u/a. That's your time, going up, with no air resistance factored in.

Coming down, as others have said, the bullet will accellerate to terminal velocity (terminal velocity v = 0 + at (u=0 for this leg) and from that you can get time to terminal t=v/a). From there it's just the terminal velocity over the remaining distance.

Of course you first have to figure out what the distance up and down is, for that you use s = ut+1/2at^2 once you have the up time t.

But as others have said air resistance will make a difference. Basically it increases a (works with gravity). But it's non-linear, based on the velocity, which is decreasing all the time. Remember, bullets have a different BC depending on the velocity.

A complex issue, but using the above equations and sticking in the MV might give you some idea.

I read somewhere that bullets fired straight up come down base first, still rotating. This is because of two reasons, firstly that the rotational velocity of the bullet is much higher than the MV and secondly that air resistance doesn't act on the rotational velocity much.

But as others have said, firing straight up is difficult and the coriolis force will also bend the flight path a bit (depending on how high the bullet goes).

OK, I'll stop now

Now I am a little confused - I thought that without air resistance, a vertically fired bullet will reach the ground at exactly the speed with which it left the ground.

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For a perfectly round object and no air resistance you'd be correct. The projectile is slowed as its height increases by the acceleration of gravity, which is opposite the direction of the velocity. At the apex of the path, the velocity is equal to zero. Now for the second leg, reverse the situation: the object falls the same distance it rose and is now sped up by the acceleration of gravity. If this occurs over the same distance as before, the velocity at the end of the path will be the same as the initial velocity of the upward leg of the path. Give me some theoretical numbers and I'd be glad to do a quick calculation.

If this occurs over the same distance as before, the velocity at the end of the path will be the same as the initial velocity of the upward leg of the path. Give me some theoretical numbers and I'd be glad to do a quick calculation.

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But, wouldn't it be possible that muzzle velocity exceeds terminal velocity?

And, so, you're suggesting that if an old rifled musket fired a perfectly smooth and round lead ball straight up, it would fall the same way (since there's no "direction" to it), but are you saying that it would reach the same ###'s coming down as when it first left the barrel? Without air resistance, how would that affect terminal velocity calculations (goes up, goes down?)

But, wouldn't it be possible that muzzle velocity exceeds terminal velocity?

And, so, you're suggesting that if an old rifled musket fired a perfectly smooth and round lead ball straight up, it would fall the same way (since there's no "direction" to it), but are you saying that it would reach the same ###'s coming down as when it first left the barrel? Without air resistance, how would that affect terminal velocity calculations (goes up, goes down?)

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Theoretically, the initial velocity just after leaving the barrel and the final velocity just before impact with the ground will be the same, in the absence of air resistance.

Adding in air resistance allows the problem to "make sense", because I'm fairly sure once air resistance is accounted for you'll see the the final velocity upon impact with the ground is much less than the initial velocity just after the bullet is fired. The problem is we'd need someone fairly well versed in calculus to do a problem like that (and my calculus is a few years rusty), as the air resistance changes depending on the velocity, which is changing because of the constant acceleration of gravity.

Bottom line:
If you ignore air resistance, final velocity is the same as initial velocity. If you include air resistance the problem gets a lot harder, but it's most likely safe to say that final velocity <<< initial velocity.

If you're given the mass of the bullet and it's initial velocity, this isn't all that hard of a problem if you ignore air resistance and only care about gravity. Someone post some fictional or real numbers and I'll do it, if you want.

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Mass has zero effect, you would not need that. If you ignore air resistance, it's a simple problem. But air resistance is there and has a massive effect. Bullets reach a terminal velocity as an effect of air resistance.

First, please see this link:

http://www.grc.nasa.gov/WWW/K-12/airplane/mofall.html

Which tells us that an object falling to Earth accelerates at 9.8 meters per second-squared.

So our hypothetical bullet, fired straight up, starts falling at 0 velocity once it reaches the apex of flight.

Applying the math (which is just straight algebra) an object falling for 30 seconds - in a vacuum - has traveled a distance of 4,410 meters and has a velocity of 294 meters per second.

At 60 seconds the object has traveled 17,640 meters (a little over 10 miles) and is moving at 588 meters per second or about 1,929 FPS.

In a vacuum a rapidly spinning object will maintain that gyroscopic spin all the way up and all the way down. So, barring any other forces, our rifle bullet will hit butt first.

To calculate Terminal Velocity in the atmosphere, see this link:
http://www.grc.nasa.gov/WWW/K-12/airplane/termv.html

Maybe we have a market for kevlar umbrellas.

Unfortunately there are lots of people out there who give handgun owners a bad name. Once in a while I will hear gunshots around 3AM. I think its some yahoos that live close to the foothills probably drunk off their ass.

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I wish that people that shoot in the air would consider shooting the gun straight down at their foot. It would really cut down on the number of shots fired at 3 AM and the number of innocent bystanders getting hurt or killed.

I have no idea how long the bullet takes to travel, but I suspect that by the time you hear the shot, there won't be much time to respond. Good luck with this. Gun shots at 3 AM can't be good.


On a somewhat related note:

Recently here in Texas, a woman was seriously hurt by round shot from 5 miles away while she was attending a NASCAR race at TMS. Granted, that was 50 caliber rifle round so it could travel that far. Fortunately, the gun owner took responsibilty for his mistake:

http://www.dallasnews.com/sharedcon...ries/110508dnmetspeedwaybullet.17df1f096.html

The preliminary ballistics tests show it was his gun:

http://www.kvue.com/news/state/stories/110608kvue_speedway_shooting-cb.188dd9855.html

I wonder how long that bullet took to travel from gun to ground...or in this case arm.