# Bullet fired from a gun and a bullet dropped descend simultaneously

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I read on quora that the bullets will hit the ground at the same time. Can anyone elaborate why this is so?

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I read on quora that the bullets will hit the ground at the same time. Can anyone elaborate why this is so?

s = v0t + 1/2 at2

What is the initial vertical velocity and vertical acceleration of each bullet?

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That is true because physical space is a three dimensional vector space. That is, we can treat different directions, in particular "horizontal" and "vertical", separately. Set up a "xyz- coordinate system". Then the acceleration due to gravity is <0, 0, -g>, acting vertically only. If the initial velociaty is <vx, vy, vz> then the velocity after time t is <vx, vy, vz- gt>. If the initial position was <x0, y0, z0> then the position after time t is <vxt+ x0, vyt+ y0, vzt- (g/2)t^2+ z0>. The time until the bullet hits the ground is given by vzt- (g/2)t^2+ z0= 0. That is independent of the x and y components of the initial position and velocity vectors.

(This has nothing to do with "relativity".)

Edited by Country Boy
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I read on quora that the bullets will hit the ground at the same time. Can anyone elaborate why when this is so?

So I have offered an amendment to your OP and title, hope you don't mind.

A bullet fired horizontally from a gun and a bullet dropped from the same height as the muzzle will reach the ground after the same journey time, ignoring the curvature of the Earth and wind resistance.

This is because both bullets start with zero vertical velocity and are only subject to the same (vertical) distance to ground and (vertical) acceleration.

Edited by studiot
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So I have offered an amendment to your OP and title, hope you don't mind.

A bullet fired horizontally from a gun and a bullet dropped from the same height as the muzzle will reach the ground after the same journey time, ignoring the curvature of the Earth and wind resistance.

This is because both bullets start with zero vertical velocity and are only subject to the same (vertical) distance to ground and (vertical) acceleration.

Studiot,

I wonder why, in the exercise, the curvature of the Earth is dropped from consideration. In the intuitive consideration of the problem, it seems to make sense, that because of the distance from the launch and the Earth curving away, the bullet will take longer to reach the ground, because in actuality the vertical trip IS longer. In fact, if you start high enough and fire a projectile horizontally to the ground the bullet will take a very long time to hit the ground, as it could continue to miss the ground and be in orbit.

Regards, TAR

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... If both bullets being compared have the same mass, volume and shape.

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So I have offered an amendment to your OP and title, hope you don't mind.

A bullet fired horizontally from a gun and a bullet dropped from the same height as the muzzle will reach the ground after the same journey time, ignoring the curvature of the Earth and wind resistance.

This is because both bullets start with zero vertical velocity and are only subject to the same (vertical) distance to ground and (vertical) acceleration.

I knew you were all clever enough to know what meant. Thanks. That makes sense.

That is true because physical space is a three dimensional vector space. That is, we can treat different directions, in particular "horizontal" and "vertical", separately. Set up a "xyz- coordinate system". Then the acceleration due to gravity is <0, 0, -g>, acting vertically only. If the initial velociaty is <vx, vy, vz> then the velocity after time t is <vx, vy, vz- gt>. If the initial position was <x0, y0, z0> then the position after time t is <vxt+ x0, vyt+ y0, vzt- (g/2)t^2+ z0>. The time until the bullet hits the ground is given by vzt- (g/2)t^2+ z0= 0. That is independent of the x and y components of the initial position and velocity vectors.

(This has nothing to do with "relativity".)

Thanks. I wasn't sure if spacetime curvature might be relevant since that is how gravity is described and I thought it might be pertinent to an accurate description of the process.

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Studiot,

I wonder why, in the exercise, the curvature of the Earth is dropped from consideration. In the intuitive consideration of the problem, it seems to make sense, that because of the distance from the launch and the Earth curving away, the bullet will take longer to reach the ground, because in actuality the vertical trip IS longer. In fact, if you start high enough and fire a projectile horizontally to the ground the bullet will take a very long time to hit the ground, as it could continue to miss the ground and be in orbit.

Regards, TAR

It is left as an exercise to calculate the muzzle velocity of the bullet necessary for it to be in orbit, briefly, before air resistance slows it down.

Suffice to say we are still trying to use reasonable, real-world numbers, even as we ignore air resistance and other confounding factors in such a problem. If it helps, you can imagine a device with a much smaller launch velocity, like a slingshot. When my high-school teacher demonstrated this in the lab, it was with a device that shot the projectile such that it hit the floor after perhaps 8-10 meters, and from a height of perhaps a meter and a half.

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Studiot,

I wonder why, in the exercise, the curvature of the Earth is dropped from consideration. In the intuitive consideration of the problem, it seems to make sense, that because of the distance from the launch and the Earth curving away, the bullet will take longer to reach the ground, because in actuality the vertical trip IS longer. In fact, if you start high enough and fire a projectile horizontally to the ground the bullet will take a very long time to hit the ground, as it could continue to miss the ground and be in orbit.

Regards, TAR

Hello, Tar

The situation of reaching orbit is unlikely as no normal gun can send a bullet at anything approaching escape velocity.

Without reaching escape velocity or having the benefit of a sustained propulsion behind it any projectile will eventually return to Earth

Wikipedia

In common usage, the initial point is on the surface of a planet or moon. On the surface of the Earth, the escape velocity is about 11.2 km/s, which is approximately 33 times the speed of sound (Mach 33) and several times the muzzle velocity of a rifle bullet (up to 1.7 km/s).

Edited by studiot
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Studiot,

I wonder why, in the exercise, the curvature of the Earth is dropped from consideration. In the intuitive consideration of the problem, it seems to make sense, that because of the distance from the launch and the Earth curving away, the bullet will take longer to reach the ground, because in actuality the vertical trip IS longer. In fact, if you start high enough and fire a projectile horizontally to the ground the bullet will take a very long time to hit the ground, as it could continue to miss the ground and be in orbit.

Regards, TAR

Because it's an illustration of the fact that lateral velocity does not directly impact vertical acceleration due to gravity. Many people are used to thinking of bullets as going "in a straight line" so they make for a dramatic example of the effect.

In the real world, obviously, variations in the terrain over distances and other factors can affect which hits the ground first, but the point of the example is that if the bullets have to fall the same distance to the ground, they will hit at the same time. If the distance changes either because of a hill, or a ditch or the curvature of the Earth or any number of other things, then you obviously won't get the same result.

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The situation of reaching orbit is unlikely as no normal gun can send a bullet at anything approaching escape velocity.

Without reaching escape velocity or having the benefit of a sustained propulsion behind it any projectile will eventually return to Earth

Not that it matters in a practical sense, but escape velocity is sqrt2 larger than orbital velocity at the surface. You "only" have to get to 7.9 km/s to orbit, and 11.2 km/s for escape. Ignoring air resistance, of course.

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Not that it matters in a practical sense, but escape velocity is sqrt2 larger than orbital velocity at the surface. You "only" have to get to 7.9 km/s to orbit, and 11.2 km/s for escape. Ignoring air resistance, of course.

Fair comment.

Thank you.

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That being said, about a month or two ago, someone posted about the longest sniper shot on record, at over two miles, where curvature of the Earth had to be accounted for. Which means there was non-negligible drop off of the Earth's surface, in which case, a longer drop, and therefore, longer time to hit the ground.

Incidentally, three of the five longest recorded sniper kills on record, are by Canadians.

( don't mess with us )

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That being said, about a month or two ago, someone posted about the longest sniper shot on record, at over two miles, where curvature of the Earth had to be accounted for. Which means there was non-negligible drop off of the Earth's surface, in which case, a longer drop, and therefore, longer time to hit the ground.

Incidentally, three of the five longest recorded sniper kills on record, are by Canadians.

( don't mess with us )

The shot took almost 10s, in which a horizontally-fired bullet would drop almost 490m, assuming no air resistance. So we can safely conclude that the shot was not horizontal, or the ground was uneven, exclusive of the curvature (i.e. shot from a cliff or tower, or similar). The normal setup of the problem would have the bullet hitting in less than a second.

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Studiot,

I wonder why, in the exercise, the curvature of the Earth is dropped from consideration.

Perhaps for the same reason we ignore the possible real-world presence of hills and walls.

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Except if you set a tangent line with a laser at 5 ft high and marked a pole a mile away (on the salt flats with no hills or walls) and put an M16 level on this tangent line, and fired it, while dropping a round from muzzle height, the round has to lose 5ft 8 inches of height to hit the ground at the pole where it loses 5 ft of height to hit the ground below the muzzle. It obviously takes longer to fall 5.67 ft than it takes to fall 5 ft.

Nothing to do with relativity yet, but gravity is gravity and curvature is curvature. If your thought experiment was on a very small spherical asteroid, and not the Earth, the tangent shot (oxygen injected to facilitate the explosion) would possibly miss the asteroid and go into elliptical orbit.

Edited by tar
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Except if you set a tangent line with a laser at 5 ft high and marked a pole a mile away (on the salt flats with no hills or walls) and put an M16 level on this tangent line, and fired it, while dropping a round from muzzle height, the round has to lose 5ft 8 inches of height to hit the ground at the pole where it loses 5 ft of height to hit the ground below the muzzle. It obviously takes longer to fall 5.67 ft than it takes to fall 5 ft.

Nothing to do with relativity yet, but gravity is gravity and curvature is curvature. If your thought experiment was on a very small spherical asteroid, and not the Earth, the tangent shot (oxygen injected to facilitate the explosion) would possibly miss the asteroid and go into elliptical orbit.

Which is why you ignore complicated set-ups when speaking about the general case and assume that extra variables are being ignored when a general statement is made.

It's the same thing as saying that two objects of different masses fall at the same rate in Earth's gravity. Yes, this is not the case if you take into account variable surface area and wind resistance, or drop the objects in the middle of a tornado, or place a strong electromagnet underneath them and make one of the objects out or iron.

Being able to recognize and accept spherical cows is an important skill.

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Well wait, my thinking is a little off, as the M16 muzzle velocity is 2900 ft per sec, and an object falls 16 ft in the first sec, so the round will hit the ground before the pole. So lets shoot the thing from a 30ft platform and have the laser mark at 30ft 8 inches. And at the pole where a tangent line has also been marked at ground level, the mark is 8 inches off the ground, so when the bullet hits the ground at the muzzle it will have fallen 30 ft at the pole, and still be traveling.

I can accept a spherical cow as a general case, but we actually live on a sphere, and if common sense would have a person think it would take an M16 round longer to hit the ground, and that is actually true, why attempt to show they are wrong by dropping out the very element that makes it true?

What is interesting is that in order to actually hit a target a mile away you have to adjust your sights so the barrel is elevated above the tangent.

you have to make a parabolic shot...a tangent shot, from eye level will hit the ground prior the mile away target, which is what the general case law is meant to illustrate

Edited by tar
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Well wait, my thinking is a little off, as the M16 muzzle velocity is 2900 ft per sec, and an object falls 16 ft in the first sec, so the round will hit the ground before the pole. So lets shoot the thing from a 30ft platform and have the laser mark at 30ft 8 inches. And at the pole where a tangent line has also been marked at ground level, the mark is 8 inches off the ground, so when the bullet hits the ground at the muzzle it will have fallen 30 ft at the pole, and still be traveling.

I can accept a spherical cow as a general case, but we actually live on a sphere, and if common sense would have a person think it would take an M16 round longer to hit the ground, and that is actually true, why attempt to show they are wrong by dropping out the very element that makes it true?

What is interesting is that in order to actually hit a target a mile away you have to adjust your sights so the barrel is elevated above the tangent.

you have to make a parabolic shot...a tangent shot, from eye level will hit the ground prior the mile away target, which is what the general case law is meant to illustrate

IOW, the problem as typically stated does not suffer to any large degree from the problems you noted.

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Here's a mythbusters demonstration of it:

Edited by StringJunky
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IOW, the problem as typically stated does not suffer to any large degree from the problems you noted.

Right. Not to any large degree. Just for answering the question yes or no as to whether the bullets hit the ground at the same time, you have to say no. Any intuitive shooter elevates above the tangent automatically, after seeing the level shot fall short, anyway. And as soon as you have elevation, you go up before you fall, and the fired bullet does actually spend more time in flight then a dropped bullet, and the firer may well consider a sight elevation "level". So right, the problem does not suffer from the problems I stated to any large degree

Regards, TAR

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Right. Not to any large degree. Just for answering the question yes or no as to whether the bullets hit the ground at the same time, you have to say no. Any intuitive shooter elevates above the tangent automatically, after seeing the level shot fall short, anyway. And as soon as you have elevation, you go up before you fall, and the fired bullet does actually spend more time in flight then a dropped bullet, and the firer may well consider a sight elevation "level". So right, the problem does not suffer from the problems I stated to any large degree

Regards, TAR

That approach (not ignoring small effects) however, leaves all physics problems as unsolvable. That saves a lot of time, but isn't particularly useful.

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Right. Not to any large degree. Just for answering the question yes or no as to whether the bullets hit the ground at the same time, you have to say no. Any intuitive shooter elevates above the tangent automatically, after seeing the level shot fall short, anyway. And as soon as you have elevation, you go up before you fall, and the fired bullet does actually spend more time in flight then a dropped bullet, and the firer may well consider a sight elevation "level". So right, the problem does not suffer from the problems I stated to any large degree

Regards, TAR

This is irrelevant to the physics I was asking about in my question,

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String Junky,

I am sorry for interrupting and deflecting the question from its purpose.

Was is more about the simultaneity of decent, where the video ignored your question by dropping the bullet where the fired bullet landed?

​Regards, TAR

As in, does the velocity cause any length contraction/time dilation?

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...

As in, does the velocity cause any length contraction/time dilation?

Any length contraction is in the direction of motion and can be dealt with in components (if you think in the frame of the pellet then there is no time dilation and all frames must give same answer) - in the y direction / vertical axis the velocity at every moment is similar for the dropped pellet and fired pellet; so no matter how small the relativistic correction might be it would still be the same for both.

And for those who quibbled that this post should not be in "Relativity" - surely this is galilean / newtonian relativity? Although forum subtitle does say it is for SR and GR

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