No, from a mathematical standpoint, that only makes sense if the drag force is directly proportional to velocity.
Here’s maybe a simpler example. Frictional force of a box sliding on the floor is constant with respect to velocity. Say it’s always 10N. So if you slide the box at 1 m/s to the east, you get 10N force to the west. If you slide the box at 1.414 m/s to the east, you still get a 10N force to the west. Say you then slide it at 1.414 m/s southeast - that is, 1 m/s south and 1 m/s east. If you could separate the frictional force into N-S and E-W components which don’t depend on each other, then you’d need 10N of friction west, and 10N of friction north, or 14.14N of friction total. See the contradiction?
I could see a possibility that the spin creates turbulence, which in some way increases the vertical drag experienced by the bullet. Stranger things have been known to happen. I haven’t seen anything treating that specifically, but rifle shooting books do mention that, in a cross wind, the spin of the bullet can have an effect, so bullets in a right-hand wind land higher, and those in a wind from the left land lower*. But if that is the only consideration, then a bullet fired from a smooth-bore musket and one dropped would land at the same time, as would one fired from a rifle and one merely spun up to 2000-odd revs per sec, and dropped.
Though it is also possible that Achernar has the right analysis. I’m not sure.
On preview, to Keeve: Yes, you would have to separate the drag into horizontal and vertical components, eventually. But the drag acts in the direction opposite to the travel, with a magnitude proportional to the square of the velocity. So first you’d have to calculate the absolute velocity and direction of travel to get the drag, which you’d then decompose into horizontal and vertical components.
E.g. ``The hunter’s guide to ballistics’’ Wayne van Zwoll. ISBN 1-58574-375-5
Q.E.D.'s answer is correct. In a vacuum, on a plane in which gravity acts equally at all points, they will hit the ground at exactly the same time. In a vacuum on a plane on the earth, the fired bullet will hit later, because the gravitational force will be less as it gets further from the center of the earth.
Once you add air resistance, it becomes more complicated, if only because fluid flow is pretty chaotic, and the kind of forces you get when moving at slow speeds from dropping as opposed to fast speeds from firing can be very different. Assuming a simpler model of particles of air colliding with the bullet, then air resistance should not make any difference, since it will always be directly against the direction of travel.
May i add a bit of a hijack here? Im a paintball player and Tippman guns have a special barrel they call a flatline barrel that is not a straight barrel but on that starts horizontally has a maybe 30-45 degree bend then goes horizontal again. Supposedly this configuration puts a backspin on the paintball and suposely gives it extra range and a flatter arc. is there any real science to this? Does backspin give extra distance?
I don’t know the technical term for this but what about ground effects, when the fired round gets very close to the ground, there may be a ‘skimming’ effect that would casue it to travel further (and stay up longer).
I do believe so. A spinning ball will alter the fluid flow around the ball, creating a lower pressure on one side than the other. Put a backspin on it, and the low pressure side is on top, which creates lift. Here’s a nifty page that explains the fluid dynamics (with an interactive Java applet! Ooooooh!)
Drag HAS to considered a force with a vector opposite direction of travel (in still air).
To keep the math easy, here is an order-of-magnitude calculation.
Gravity is 10.0 m/s.
Bullet dropped is going 10.0 m/s at t = one second.
Drag (f[sub]d[/sub]is roughly square of velocity, or 100. Vertical component of that (d[sub]vert[/sub]) is 100 (all of it).
Bullet fired at 500 m/s. It has a horizontal velocity (v[sub]h[/sub]) of 300 m/s at t = one second. We’ll say it also has a vertical velocity (v[sub]v[/sub]) of 10m/s at t = one second.
Even in a vacuum on a plane where the force of gravity works the same at all points they would fall at different speeds… Time would slow down ever so slightly for the fired bullet (or speed up, i forget which) thus producing the very very very slight difference.
Yup, Uncle Bill is right – if drag is proportional to the square of the speed and opposite to the velocity vector, then adding horizontal speed increases vertical drag. Of course, air resistance in the real world is much much much more complicated that this simple model, so the real answer depends on the shape of the bullet, its spin, etc.
And for brainfizz, if you’re going to bring relativity into it, then the answer is that different observers in different frames of reference (i.e. moving at different speeds) will disagree on whether the bullets hit simultaneously, so there is no one right answer.
(remember, in a vacuum with an infinite flat featureless plain, there’s no way to tell which bullet is moving… so time has to slow down for each of them, when you look at it from the other bullet).
Yes, it would. This phenomenon is easily observed in golf balls, whose backspin increases their distance of flight enormously, or in baseball; a pticher throwing a particularly hard fastball puts backspin on it, preventing the ball from dropping as much as it would otherwise on its way to the plate
Anyone who’s played paintball can attest to the fact that paintballs, like golf balls or baseballs, are extremely prone to curving in the air due to their spin.
I think the gravitational differences between shooting a gun on a plane vs a sphere would be too small to measure. But there is a measurable distance caused by the fact that the ground is parallel to the flight path on a plane and “falls away” from the flight path on a sphere. The effect is noticable that it has to be compensated for in long range naval gunfire.
As long as we’re mixing firearms and relativity, let’s consider the fact that the shot bullet, due to its higher velocity, will have a greater mass than the dropped bullet. Hence the more massive shot bullet will be attracted more by the earth’s gravity and will fall faster and land sooner.
And to expand on my 11:20 post, the curvature of the earth is approximately eight inches per mile. Most firearms have a maximum range of over a mile, so the effect will be that one bullet is falling sixty inchs while the other has to fall over sixty eight.
Here is a trajectory calculator. Notice that as you increase the muzzle velocity the bullet travels further before hitting the ground but the time taken to hit the ground remains the same.
Let’s just say they would essentially hit the ground at the same time. At exactly the same time, probably not - for the myriad of reasons posted here. My physics professor in high school performed an experiment for us to prove this point. Let me see if I can explain what he did…
He had one bullet (well, in this case it was a tennis ball) suspended from the ceiling on one side of the classroom and another tennis ball loaded in a homemade cannon device suspended from the ceiling on the other side of the room. Both tennis balls were at equal elevations from the floor, the cannon was level and aimed directly at the tennis ball suspended from the ceiling. He had a device rigged so that when he fired the tennis ball from the cannon it would trigger a switch on the end of the muzzle that had a wire connected to a device holding the other tennis ball from the ceiling. So what this system would do is release the stationary tennis ball as soon as the ball from the cannon left the mussle. When he shot the tennis ball from the cannon it flied across the room in its predicted parabolic fashion and collided with the other tennis ball in mid air. So, what he proved was, no matter how fast the ball from the cannon was shot, both tennis balls were falling at the same velocity and they would both reach the ground at essentially the same time.