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If you have a gun which is pointing horizontally. And shoot it, you also drop a bullet from exactly the same height. What hit s the floors first.
Please if you can include reasons for your answers
Same time.
Won't the curvature of the earth's surface mean that the bullet fired from the gun will have further to go to reach the ground, and so land later?
depends on the range of the gun I'd have thought
dropping bullet horizontally would have much slower acceleration obviously, however would have no horizontal motion
so in other words, I don't really know 🙂
The same time...
Although it could be argued that the curvature of the earth could have some impact, but the amount of the curve over the distance that a bullet travels is insignificant.
Not sure richmars is taking into account the curvature of the earth.
been done on mythbusters.
[url=
bullets video[/url]
same time by the way
In a vacuum then yes they would. The force acting on both bullets in the vertical plane is gravity (assuming a vacuum) so they'd hit the floor at the same time.
In air, it'll depend if the one you drop from your hand naturally falls nose first - then it'll have a slightly lower coefficent of friction than the spin stabilised gun bullet.
If the gun is pointing horizontally then the bullet will travel in an arc.
So the bullet dropped from the same height will hit the floor first as its velocity is in only one direction whereas gravity has to overcome the lateral velocity of the bullet.
(Disclaimer - I haven't a clue to be fair and am blagging it going on the long range rifle shooting I did many moons ago.)
Is the gun on a conveyor belt?
No, I didn't. But then the ground isn't perfectly flat so could just as well be slightly higher as lower where the bullet lands.
Approximating the surface of the earth to be locally flat, they will land at the same time.
The downward acceleration on each bullet is the same: 9.81 m/s.
This simple solution ignored boundary effects of the spinning bullet but you're not paying me to think about that.
My old physics prof, when talking about this problem (bullet fired at villain hanging off a ledge, who drops at the same moment the bullet is fired - what happen) described it as "ah, it's the old monkey and the hunter". We pissed ourselves laughing at that for hours 😆
you have to look at the motion in two directions, x and y. one bullet has horizontal motion when it leaves the barrell the other does not when it is dropped. Both have no vertical motion initially.
If you now consider the forces acting on the bullet. horizontal motion is affected by air resistance, so the moving bullet will slow. both bullets experience the same vertical force acceleration, gravity.
Who watched QI last night then.
[i]gravity has to overcome the lateral velocity of the bullet[/i]
No it doesn't, and that's the whole point.
It's a purely theoretical point about force and mass and ballistics, but perfectly sound physics. Since it's theoretical in a certain domain you need to make a few assumptions: the ground is flat, and you ignore fluid dynamics.
Both bullets start with a vertical velocity of zero. Gravity acts vertically and accelerates both bullets vertically at the same rate. After one second, both bullets are moving vertically at exactly the same velocity.
The fact that one bullet has a high horizontal velocity and the other zero is irrelevant to the vertical velocity.
The point it demonstrates is that a force affects the velocity of a mass in the direction of the force only. The force has no effect on any perpendicular velocity, and vice versa.
In the words of Scroobius Pip 'Thou shalt not question Stephen Fry'
If I were to fire an aeroplane from a suitable gun, & simply drop one from the same height with no forward motion, will they still hit the ground together...
Will the forward motion not serve to increase / decrease the density of the air under / on top of the wings & hence induce some lift...?
I guess the bullet's shape wont be subject to the above so consider takisawa2 thick.
[i]If I were to fire an aeroplane from a suitable gun, & simply drop one from the same height with no forward motion, will they still hit the ground together...[/i]
No, because as you rightly say, forward motion will generate lift. That's why it's a simple force/mass example and not a fluid dynamics example.
If I were to fire an aeroplane from a suitable gun,
Go on then...
[i]...it'll have a slightly lower coefficent of friction...[/i]
Coefficient of friction is a material property and is completely unchanged. You (hopefully) mean coefficient of drag.
is the gun on a conveyor belt?
If the gun was big enough, the shell might never land. Conveyor belt or not.
What if the gun was fired by an octopus?
Fire it fast enough and it'll never hit the ground...
Forward motion alone won't be enough for a bullet to generate lift. It would require a shape such that the pressure distributions above and below it are different[1]. Either that, or it would need some spin (pitching plane, not axial plane (rifle-tastic)).
References [1] JD anderson - fundamentals of aerodynamics
If you had totally flat terrain and you're ignoring the curvature of the earth, the dropped bullet will still hit the ground first. That's because I refuse to ignore fluid dynamics - I'm just going to simplify it somewhat. The horizontal motion of the bullet does have an effect on the vertical motion because aerodynamic drag force is non linear (is proportional to the square of the speed) - this means that there's more vertical drag force on the fired bullet.
Never mind bullets, am I the only one seriously impressed with the guy on the motorised unicycle ?
Horizontal motion does not have an effect, fluid dynamics or not.
For a start, bullets are symmetrical. Even symetrical aerofoils don't generate lift at zero incidence.
Please don't simplify, indulge us.
Ewan's got it.
For the OP...i have no idea at all...my mind says shot. I cannot for the life of me see how it's anything else...but my science is balls.
Sammysammsamm wrote,
"Forward motion alone won't be enough for a bullet to generate lift. It would require a shape such that the pressure distributions above and below it are different[1]. Either that, or it would need some spin (pitching plane, not axial plane (rifle-tastic))."
It doesn't need to generate lift, it just needs sufficient velocity.
Horizontal motion does not have an effect, fluid dynamics or not.
Well yes it does - for the reason I gave, that the drag force is non linear (proportional to the square of the speed). It doesn't need to generate lift, just drag.
To give some numbers, this means that if the bullet is going at 10m/s vertically (after just over a second of falling), then the vertical drag force on the dropped bullet is 100N, where N is a constant.
If the bullet is doing 100m/s horizontally (it will be going far faster when fired, I'm trying to make numbers simpler), then when it is also doing 10m/s vertically, the total speed is 100.5m/s. This results in a total drag force of 10100N in the opposite direction to the direction the bullet is moving in, the vertical component of which is 1005N. So 10 times the vertical drag force.
You're in radians, it's late and I'm unsure if I should be too.
Northwind - I was agreeing with you! I hadn't read your post 😀
I was talking to Bez
I retract that, your maths is sound - Watch this space.
I did once calculate how fast the bullet would need to be going to never fall any closer to the centre of the earth. i.e. it's rate of fall is the same as the curvature of the earth. i.e. it's in orbit.
if you're going to bring drag into the equation it's much more complex than aracer is putting it. rifling, golf ball surfaces and spin, etc. etc.
Theoretically at the same time as the vertical component of acceleration, and therefore velocity (as they both start from 0) will be 9.81m/s.
(This is not taking into account the curvature of the earth etc although I don't think the range of the gun is big enough for that to make much of an effect anyway)
For all intents and purposes (ie, a normal gun - pistol or rifle, no atomic artillery) then the same time. As others have said, the earth is near enough flat over the kind of distances we're talking.
Start getting silly and yes, you could fire an item into orbit from a gun, given a big enough gun.
Either my eyesight deceives me or none of the STW pedants have got here yet...
The downward acceleration on each bullet is the same: 9.81 m/s
That'll be 9.81m/s^2, good sir 🙂
That'll be 9.81m/s^2, good sir
my excuse is I'm a lazy chemist and not a physicist 😉
I do this is a demo. I fire a ball bearing across the lab. As it leaves the "gun" it tears a piece of foil. This turns off an electro magent releasing a can which is at the same height as the gun. The can and ball bearing both fall about .7m before colliding (on a good day) with an audible ping.
Its a great fun demo and I love the debate before hand about whats going to happen. We usaully do it as the monkey hunter and the monkey. The monkey lets go of the tree as the hunter pulls the trigger
I think my demo uses low velocities so we can ignore air resistance
I think my demo uses low velocities so we can ignore air resistance
Well that's where you're going wrong - if you use a real gun with normal muzzle velocity you can't ignore air resistance 😉
This has been done by Mythbusters and I think the answer was that they hit at about the same time.
A bullet won't generate lift due to it's shape since they tend to be symmetrical for obvious reasons. Spin around the (horizontal) axis of travel won't generate lift either.
Assuming a flat earth, the only thing that might make then not hit the ground at the same time are:
a) if the dropped bullet doesn't stay horizontal, so it has a very slightly lower air resistance vertically (unlikely to make much difference).
b) Relativistic effects on time (unlikely to be even measurable for bullet speeds).
Ah yes, aracer - good physics.
He's right, and that's also the reason why a side wind whilst driving will have a negative impact on fuel economy (I think).
I 'spose the nearest you can get this scenario to a conveyor belt is a geostationary satellite. It's moving really fast but it's not...freaky!!
Doesn't a bullet decrease in mass once fired? That would mean the stationary bullet would accelerate quicker and hit the ground first.
aracer - MemberHorizontal motion does not have an effect, fluid dynamics or not.
Well yes it does - for the reason I gave, that the drag force is non linear (proportional to the square of the speed). It doesn't need to generate lift, just drag.
To give some numbers, this means that if the bullet is going at 10m/s vertically (after just over a second of falling), then the vertical drag force on the dropped bullet is 100N, where N is a constant.
If the bullet is doing 100m/s horizontally (it will be going far faster when fired, I'm trying to make numbers simpler), then when it is also doing 10m/s vertically, the total speed is 100.5m/s. This results in a total drag force of 10100N in the opposite direction to the direction the bullet is moving in, the vertical component of which is 1005N. So 10 times the vertical drag force.
Am not sure how you got 1005N out of that... just remember similar triangles and all that...
If the horizontal velocity is 100m/s and the vertical velocity is 10m/s then you are correct that the overall velocity of the bullet is 100.5
If the drag works on the square of velocity, then you can do the calcs on it's overall velocity, but you can also do it on the vertical and horizontal components, and it will come to the same answer.
So, if you square the overall velocity to get your overall drag, so 10100N. You can also do it by squaring the horizontal velocity and adding it to the square of the vertical... (10000 + 100)N
It doesn't change the vertical component at all, which stays the same as the bullet that has been dropped.
shaggy... tut tut.. didn't your physics teacher teach you anything? Irrespective of mass, both bullets would fall at the same rate... I think Galileo proved this a few years ago...
:o)