Take h and v to be the horizontal and vertical (v points down) unit vectors respectively, and Vhat to be the velocity unit vector. (I’m assuming Chronos’s statement that “air resistance is proportional to speed squared” is correct for the relevant velocities.)
So, let’s see if I’ve got this right. If it takes a bullet 1 second to hit the ground from point x, and a bullet fired from the gun we are using, at same point x, travels at 300m per sec, I only have to stand 301 m from the firee to see the bullet drop harmlessly before me?
Sorta. I guess if you’re firing it at abour five feet, the dropped bulled will hit the ground in about a half second. That’s about 150 feet for the fired bullet. But then in that half second, the fired bullet should also slow down somewhat, so it might be less than 150 feet traveled.
Only if the gun is fired exactly horizontally. I think this is why there are adjustable sights on guns, to compensate for bullet drop at longer ranges.
In the ideal scenario, yes, although of course it won’t travel 300m horizontally, then abruptly stop and fall to the ground - I don’t think you were expecting that it would, but some readings of your post could be taken that way.
But wouldn’t it be cool if it did? ( I do realise it is going to follow some sort of arc, before dropping at my feet. ) Thanks for the explanations, everyone.
**The dropped bullet will hit the ground before the bullet fired from the gun.
For a cite, check out (the very authoritative) *Understanding Firearm Ballistics *by Robert A. Rinker.
I have the book right here in my hands. On pages 164 through 165, Mr. Rinker addresses this very topic:
(FYI, the above excerpt is permitted according to the book’s copyright statement.)
The thing you’re all missing is that a bullet has a nose high trajectory which creates aerodynamic lift. This is not common knowledge, even to seasoned shooters, so don’t feel bad if you missed this crucial detail.
IANAShooter, but I fail to see why a bullet, which in my limited experience is symetrical around a horizontal axis, would have a “nose high” trajectory. Does that mean it is not travelling horizontally? Is this anything to do with a spin imparted by a rifled barrel?
Musicat, here’s a description I found. In short, gyroscopic effects due to the bullet’s rotation about its axis, and the change of that axis as the bullet follows its arcing trajectory, keep the bullet nose higher than it would otherwise be.
Well, Chronos already gave the same answer.
From the description in the above link, “nose high” seems to mean high relative to the direction of travel, rather than relative to horizontal. The bullet would have to have some"nose height" to compensate for the arc it’s following just to be level during its flight, rather than angling down.
He’s right, of course, that the orientation of the bullets matters, but this “nose high” behavior may just mean that the bullet is angled less downward, not actually angled upward. The difference in time to impact in the book (Is that a theoretical result? Do they show their calculations?) could be attributable to the difference in drag Chronus pointed out.
The book spends many pages talking about yaw, precession, gyroscopic drift, etc. The subject matter is *quite *complex. If you’re interested in this subject, I highly recommend getting the book.
One thing is not obvious to me perhaps due to my limited experience with guns: are we talking about a spherical bullet or one shaped like a rocket, i.e., pointed at the forward end? It seems that a sphere would have different flying characteristics than a pointy-shaped missle, especially if the possibility of the missle tumbling is considered. If it’s a sphere, I don’t see how we can use the term “nose high”, as it doesn’t have a nose. But I can see a spin imparting some influence on the trajectory like a curve ball works in baseball. Then the question is does the influence cause the bullet to drop, rise, or go in some other direction? I thought the reason a curve ball curves is because the axis of rotation is vertical, and one side of the ball has increased drag, the other side, lesser. In the bullet’s case, rifling on the barrel would make it spin on a horizontal axis, and there would be no increased drag in any direction.
Here’s what common sense tells me; if bullet A is dropped, and bullet B is fired simultaneously, bullet B will be travelling along a horizontal line at least until gravity overcomes the velocity of the gun, and it therefore has no chance of overtaking bullet A, which has been dropping since it was let go, with no other forces acting on it. From what I’ve read above, this seems to confirm my common sense feeling. Or am I reading the replies wrongly?
Gravity acts on both bullets the same way. It acts on the fired bullet the second it leaves the gun barrel. In the absence of air, both fall at the identical rate. Cite.
Common sense may tell you otherwise – and common sense is often wrong – but gravity doesn’t wait to act just because a body has a rapid forward motion. It’s there all the time.
If you were to ignore the forward component of the fired bullet, and charted its vertical motion vs. time, this chart would be identical to the dropped one, in a vacuum, of course. Both fall the same distance in the same time.
But it obviously doesn’t! When the bullet that is fired leaves the barrel of the gun, it doesn’t begin to move downwards unless it is pointed in that direction. Are you saying there is no maintaining of a specific height before the bullet starts its downward arc?
One post above, with a reference, seems to be telling me that my assessment is correct, and that the fired bullet does indeed take longer to hit the ground.
In a non-vacuum, gravity is still interacting with it, the AIR and motion (spin?) of the fired bullet is just messing with it.
In Physics-Land where everything is perfect, nothing can go wrong, that pesky air doesn’t bother us and the bullet moves straight, without turning at all, the bullet will start moving down the instant it leaves the barrel. Going forward has no effect on it going down, they’re two completely different forces. The sum of every force acting on the bullet is effecting what it does. In this case its weight (mass times acceleration due to gravity), a perfectly vertical force, and the propulsion due to the explosion, a perfectly horizontal force (again, Physics Land) add up to make it curve downward (curve because it’s accelerating downward, not instantly going at max speed).
In other words any random force has no bearing on any other force, it’s simply what the numbers add up to in each direction that gives you your result. What we’re doing here is simply putting in more things to add up and seeing if the numbers come out differently in a more complex situation, which so far it seems that they do.
This is a good online textbook that’s really easy to understand, I just linked to the simple 2D Newtonian Mechanics portion, lesson two is the one of interest here, read lesson one if you don’t understand vectors (it’s not a difficult concept, I promise).
( I do realise it is going to follow some sort of arc, before dropping at my feet. ) Thanks for the explanations, everyone.