Drop a bullet vs. Shoot a bullet

[91TTZ]

Originally posted by: Connoisseur
Since the earth is so big for all practical purposes any point on the earth is virtually flat. so from our perspective they would both hit the earth at the same time. however, the fired bullet would hit the earth a few milli/micro seconds later. /end thread

P.S. I didn't read all of the threads

If you did, you'd see that I worked out that the shot bullet would need to fall an extra couple of inches.

 

[akubi]

Originally posted by: Bigsm00th

Originally posted by: akubi

oh? and when did you prove your first equation v_f^2=v_i^2..blahblah?

in fact, my reasoning is more fundamental than your convoluted approach.
any object with acceleration a moves (1/2)at^2 (basically derives from the definition of acceleration)
so plug in g for a and 1m for x and solve for t. what more is there to say? your equation derives from a set of these fundamental equations.

you are asking me to prove equations derived by centuries of physicists. i will ask you in the same way to prove what you are saying so you can see how stupid it is.

again, you completely missed the point. what you just said is elementary to me but thats because it is my area of interest/study. others may not see it that way. you are being stubborn and trying to flex you e-muscles by "calling me out", even though you know my explanation is correct.

oh it's simple. velocity is defined as the rate of change of distance travelled with respect to time. acceleration is rate of change of velocity with respect to time.
so acceleration is the second time derivative of distance. integrate twice and there you have it.

 

[MrDudeMan]

Originally posted by: akubi

Originally posted by: Bigsm00th

Originally posted by: akubi

oh? and when did you prove your first equation v_f^2=v_i^2..blahblah?

in fact, my reasoning is more fundamental than your convoluted approach.
any object with acceleration a moves (1/2)at^2 (basically derives from the definition of acceleration)
so plug in g for a and 1m for x and solve for t. what more is there to say? your equation derives from a set of these fundamental equations.

you are asking me to prove equations derived by centuries of physicists. i will ask you in the same way to prove what you are saying so you can see how stupid it is.

again, you completely missed the point. what you just said is elementary to me but thats because it is my area of interest/study. others may not see it that way. you are being stubborn and trying to flex you e-muscles by "calling me out", even though you know my explanation is correct.

oh it's simple. velocity is defined as the rate of change of distance travelled with respect to time. acceleration is rate of change of velocity with respect to time.
so acceleration is the second time derivative of distance. integrate twice and there you have it.

:roll: thanks ive been through calculus. this isnt even worth my time. you didnt even prove what i asked you to. go ahead and continue to call me whatever you want, im checking out of this thread.

 

[illusion88]

none of you are adding in the time it takes for the bullet to leave the barrel. So actually the bullet dropped would land first, because the bullet in the gun didnt start to drop until it left the barrel. The time difference is probably measured in miliseconds.

 

[GreenGhost]

After reading this thread, I now believe the average age of ATOT members is 5.5 years.

 

[zerocool1]

Originally posted by: OffTopic

Originally posted by: Koenigsegg
Same?

They both feel the same down force exert on them therefore they both drop at the same rate.

yep....

 

[NissanGurl]

Originally posted by: gigapet
barring no air resistance and gravity remains constant they hit the ground at the exact same time.

Horizontal acceleration is independent from vertical acceleration, so the above is my answer. Do we all get cookies now?

 

[Armitage]

Originally posted by: NissanGurl

Originally posted by: gigapet
barring no air resistance and gravity remains constant they hit the ground at the exact same time.

Horizontal acceleration is independent from vertical acceleration, so the above is my answer. Do we all get cookies now?

Only if the earth is flat.

 

[dxkj]

Originally posted by: mchammer187

Originally posted by: 91TTZ

Originally posted by: LordMorpheus


no, you can't. A gun works by creating a spark that ignites the gunpowder . . . the sudden expansion due to the burning poweder pushes the bullet out of the barrel . . .

However, gunpowder won't burn unless there is oxygen available for the reaction. Go on, take a gun into space and pull the trigger. You'll get nothing.

Dead wrong.

Gunpowder doesn't need oxygen, as it's comprised of both the fuel AND oxidizer.

we are not talking about zero gravity we are talking about a vacuum which relies on the expansion of gases in the chamber to propel the bullet forward

if there is no gases to expand there is no propulsion

this is the reason why you can't shoot guns underwater or why you cant start a fire underwater

are you joking with me?? the vacuum becomes a non-vaccuum when the gun fires and adds whatever is produced from combusting gun powder

 

[bigredguy]

Since this is on earth somewhere, there are numemous factors that come into play. On some flat plane in a vacuum and so gravitational pull it should be the same. But i belive variations in wing, air temp, air press and millions of other little details would cause micro-variations in the way the bullets dropped. And that the bullet that was fired will be slight different shape and temp then the uncased, unfired bullet, there are so many differences, i would say that if tested in real life there would be very instances where they hit even close to the same time.

Assuming that the fired bullet was the exact same as the unfired and perfect conditions, they should hit the same time.

 

[Aquila76]

Originally posted by: bigredguy
Since this is on earth somewhere, there are numemous factors that come into play. On some flat plane in a vacuum and so gravitational pull it should be the same. But i belive variations in wing, air temp, air press and millions of other little details would cause micro-variations in the way the bullets dropped. And that the bullet that was fired will be slight different shape and temp then the uncased, unfired bullet, there are so many differences, i would say that if tested in real life there would be very instances where they hit even close to the same time.

Assuming that the fired bullet was the exact same as the unfired and perfect conditions, they should hit the same time.

Chaos Theory would also show that the results could never be duplicated because of all those little variations (wind speed/direction, air temp/pressure, humidity, finger temp/sweat/grip, etc.), so the results would always vary - even if done in the same place by the same person.

That being said, and in line with your sig, WHO FVCKING CARES?!?!?

 

[Cattlegod]

Originally posted by: akubi
well if you consider the curvature of the Earth and its rotation, the answer isn't as trivial as you might think

exactly

 

[Cattlegod]

you guys are also forgetting one thing.

velocity changes time. time would actually slow down for the bullet as well. They won't hit the ground at the same time - the fired one falls further. also more time passes for the dropped bullet because it's velocity is slower.

This means that the difference from the earth's curve will not make as much of a difference because time will move slower for the dropped bullet.

 

[EatSpam]

Originally posted by: gigapet
barring no air resistance and gravity remains constant they hit the ground at the exact same time.

 

[HardcoreRobot]

Originally posted by: Cattlegod
you guys are also forgetting one thing.

velocity changes time. time would actually slow down for the bullet as well. They won't hit the ground at the same time - the fired one falls further. also more time passes for the dropped bullet because it's velocity is slower.

This means that the difference from the earth's curve will not make as much of a difference because time will move slower for the dropped bullet.

irrelevant because the time is constant as observed by a third party (it might seem longer for the bullet, but thats really inconsequential)

 

[QuitBanningMe]

No curvature = same time
Add curvature and a long distance = the dropped bullet
Add enough velocity to the fired bullet and it won't hit the ground

 

[OutHouse]

Originally posted by: RedRooster
From 4' high you drop a bullet(without casing) and at the same moment you fire a gun exactly parallel to and 4' from the ground. Which bullet would hit the ground first?

edit- ok, you walk out in the middle of a field on a normal day(not in space, not in a vaccum) and try this. standard .22 calibre rifle. and we can assume the field is perfectly flat, with no hills of dirt sticking up, you get the idea. or it could be all cemented, with hundreds of yards of cement(or however far a .22 will shoot) EXACTLY flat the whole way, not curving with the Earth's surface along the way.

exactly Parallel? so you shoot a bullet into the ground as you drop one? well the fired bullet will hit first.

 

[QuitBanningMe]

Originally posted by: Citrix

Originally posted by: RedRooster
From 4' high you drop a bullet(without casing) and at the same moment you fire a gun exactly parallel to and 4' from the ground. Which bullet would hit the ground first?

edit- ok, you walk out in the middle of a field on a normal day(not in space, not in a vaccum) and try this. standard .22 calibre rifle. and we can assume the field is perfectly flat, with no hills of dirt sticking up, you get the idea. or it could be all cemented, with hundreds of yards of cement(or however far a .22 will shoot) EXACTLY flat the whole way, not curving with the Earth's surface along the way.

exactly Parallel? so you shoot a bullet into the ground as you drop one? well the fired bullet will hit first.

 

[hypn0tik]

Originally posted by: Cattlegod
you guys are also forgetting one thing.

velocity changes time. time would actually slow down for the bullet as well. They won't hit the ground at the same time - the fired one falls further. also more time passes for the dropped bullet because it's velocity is slower.

This means that the difference from the earth's curve will not make as much of a difference because time will move slower for the dropped bullet.

Irrelevant as it only applies to relativistic speeds. If a bullet were to approach such speeds, it'd be long gone and orbiting earth.

 

[Cattlegod]

Originally posted by: hypn0tik

Originally posted by: Cattlegod
you guys are also forgetting one thing.

velocity changes time. time would actually slow down for the bullet as well. They won't hit the ground at the same time - the fired one falls further. also more time passes for the dropped bullet because it's velocity is slower.

This means that the difference from the earth's curve will not make as much of a difference because time will move slower for the dropped bullet.

Irrelevant as it only applies to relativistic speeds. If a bullet were to approach such speeds, it'd be long gone and orbiting earth.

the effects are still there, although they are small at any speed.
the only difference it makes is each bullet and the 3rd party observer would measure different times when they all hit.

 

[chrisms]

For simple math purposes they hit the ground at the same time, but if done in the real world (still assuming a perfectly flat surface and a shot parallel to the ground) numerous other factors would come in that I don't want to even bother messing with.

 

[g8wayrebel]

Originally posted by: akubi
well if you consider the curvature of the Earth and its rotation, the answer isn't as trivial as you might think

Gravity is effective relative to the earth , so the curvature is irrelevant. They both hit the ground at the exact same time if no environmental aspect affects the fired round.

BTW... satellites do fall(Skylab ring a bell?). The thrusts are used to keep them in the desired orbit as well as defeat the pull of gravity. There is no resistance to motion in outer space , only the gravitational pull of the nearest celestial object.

 

[sciencewhiz]

Originally posted by: hypn0tik
Irrelevant as it only applies to relativistic speeds. If a bullet were to approach such speeds, it'd be long gone and orbiting earth.

If a bullet were to approach relativistic speeds, it would be long gone from earth orbit too.

 

[akubi]

Originally posted by: g8wayrebel

Originally posted by: akubi
well if you consider the curvature of the Earth and its rotation, the answer isn't as trivial as you might think

Gravity is effective relative to the earth , so the curvature is irrelevant. They both hit the gorung at the exact same time if no environmental aspect affects the fired round.

you're funny

 

[Squisher]

I had to look at this thread after seeing it on the top page for two days.

I can understand people not realizing all the factor that affect a bullet shot out of a gun, but those that are wrong seem excessively staunch in their views.


BTW-there will be some air inside the shell of a bullet taken into a vacuum and if there wasn't it still would ignite. (how does solid rocket propellant work?) Curvature of the earth would come into play.
I wanna gun that fires a bullet at the speed of light. I just wouldn't want to hold it and fire it.