Your understanding is exactly wrong. The bullet does begin to move downwards as soon as it leaves the gun barrel. There is no maintaining of a specific height before the bullet starts its downward arc.
There’s a physics demonstration generally called the “monkey gun” to show this. The idea is that there is a monkey hanging from a tree, who let’s go and drops at exactly the time you fire your gun. Where should you aim? The answer is you aim directly at him(*), and it doesn’t matter how fast the bullet goes, it will hit him.
There’s a bunch of YouTube monkey gun videos, but I don’t have the patience to find a good one. Instead, here’s a strobed picture that makes it clear what is happening. In that picture, if the woman had blown harder, the bullet would arc less, and hit the target higher up. But no matter how hard she blows, it will always hit the target (well, except that the floor will get in the way if she doesn’t blow hard enough.)
(*) This isn’t a real gun where the sight is adjusted to correct for the bullet dropping with distance, it’s a monkey gun. The axis of the barrel is aimed directly at the monkey.
Forgot about this. For the talk about “nose high” trajectory, the bullet is, well, bullet shaped (part 1 in the picture). Most of the talk has been ignoring the shape (or implicitly assuming spherical).
Your statement is correct when the bullet leaves the barrel, but as it starts to fall, the axis is no longer exactly parallel to the direction of travel, and you get some nonsymmetric drag effects. For a bullet-shaped bullet, that can affect the up/down trajectory. For a spherical bullet, the spin axis would always be parallel to the plane the bullet is arcing in, but it could give the bullet some small left/right effect (which means the bullet isn’t quite traveling in a plane, and there’d be some really small up/down effect).
Alright, everyone here who’s a toon needs to come clean.
Been thumbing through Rinker’s book a little more. Apparently the U.S. Army did some penetration tests a few decades ago, and discovered a .30-caliber bullet has more penetration at 200 yards than 100 yards. It was due to yaw… as the bullet leaves the barrel, there is a significant amount of “nose-high” yaw. At a very close range, penetration is inhibited by this yaw. (When the bullet hits the target at close range, the bullet’s longitudinal axis is not parallel with the velocity vector, thereby inhibiting penetration.) The yaw angle decreases at longer distances, thereby improving penetration. But penetration will eventually decrease at a certain distance due to loss in velocity.
Assuming the fired bullet is fired from a gun in an exactly horizontal direction, yes; the instant the bullet leaves the gun barrel it begins to fall, and at the same rate as a bullet dropped without any forward motion. The forward motion has no effect on the vertical motion and vice versa.
This is elementary high school physics, at least as I remember it. If it seems illogical to you, perhaps a refresher course in Newtonian mechanics would be helpful:
You know, thinking of this, the length of the barrel could have some (minor) bearing on the problem, couldn’t it? It would probably be negligible, but firing a bullet from, say, a rifle with a longer barrel could making it take ever so sliiiiiiiiightly longer to hit the ground than the dropped one, though I’m not sure if it’s even enough time to make it worth considering.
Ha! Yes, that occurred to me, as the bullet won’t start dropping until leaving the barrel, as the barrel is holding it up until then. It would certainly be a very, very short time, but if you weathered the “plane on a treadmill” discussions, you know that such insignificant factors, to some people, must not be ignored.
It doesn’t even have to be simultaneous, Zen. You pick a pistol with a suitable range - 100 m, say, you have some way of registering when the bullet lands at the extent of its velocity. You fire the gun at a fixed height and point, then time how long it takes to land. Then, you drop the same calibre bullet fom the same point and time that. The difference might be hardly noticeable to the human eye, but I’d wager a significant amount - if I had it - that the dropped one would land first, and there are 2 replies on here that seem to agree.
The latter. I’m not even convinced Cecil explains the original question that well in his addressing of it. Someone is going to have to walk me through this…if they have the patience.
Since when did they( the underlined) become scientific terms? Aren’t planets moving objects, and don’t they travel in arcs? And how many bullets travel far enough to be affected by the curvature of the earth?
And the last bit is an irrelevant aside. How conclusive is that?
Yes, you’re correct. Assuming the 300 m/s muzzle velocity* that some quoted earlier, in a pistol with a 4 inch barrel (about .1 m) the bullet takes about 1/3000 of a second to exit the barrel, while in a rifle with a .6 meter barrel, the bullet takes 1/500 of a second to cover the same distance.
*Which we can’t, really, because the bullet actually accelerates through the barrel, and this royally screws us up because the bullet will then need to accelerate slower through the long rifle barrel to reach the same muzzle velocity, but just for the sake of exercise…
The length of the barrel is irrelevant; we are timing it from the point it leaves the barrel, aren’t we? After all, that is the first opportunity gravity has to act on it?
The last bit is how Isaac Newton came up with the idea that gravity could explain how the moon orbits the earth. The classic Newtonian thought-experiment is to imagine being able to throw a rock as fast as you want to, and what the effects would be. It’s the logical extension of discussing that the curve of the earth affects the problem.
Let’s forget about outer-space. Say we had a warehouse sized vaccuum on Earth, and we did the experiment outlined above; what is going to happen? Forget about barrel sizes and magical orbiting bullets, just someone please simplify all this for me.
Actually, the space part makes it easier. If you have a magical gun that can propel a bullet to a magical speed it will never hit the ground, right? Well if you logically slow that down it will be at a different part of the curve (instead of missing it altogether).
Out of wonder, do you happen to have a Graphing Calculator of any kind? I have a parametric equation (One for bullet 1, one for bullet 2, and one for the Earth, in a vacuum of course) and window setting I can use to show you (I was bored in Trig one day a couple years ago)
I think this might be one of those instances when the Dope should allow images. The graphing calculator idea sounds scary for someone with my math ability…
eta. p.s. Am I “physics dumb”, or is this supposed to be hard to get your head round?
If you remove air from the situation, the bullet fired from the gun and the bullet dropped from the same height will hit the ground exactly at the same time. In the absence of aerodynamic effects the horizontal velocity of the bullet has no effect on the rate of fall. That is determined only by gravity, and effects all objects identically.
Only when you add air and the resulting aerodynamic effects does the forward velocity begin to have an impact on the rate of fall.
And the curvature of the Earth, which is what we were discussing.
Ivan - the graphing calculator isn’t hard, I’d even tell you what to type in. Anyway did you read the links to the textbook I gave you? That should clear everything up, it’s really simple once you read it, and the site doesn’t have any viruses, I promise.
My teacher did a test for me a while ago. Granted it was tennis balls (bullets are hard to follow, surprise) but they hit at the same time (so long as released both at the same time…)
Yes, that occurred to me, as the bullet won’t start dropping until leaving the barrel, as the barrel is holding it up until then. It would certainly be a very, very short time, but if you weathered the “plane on a treadmill” discussions, you know that such insignificant factors, to some people, must not be ignored.