Ok, I’m in an airplane, say a 747 cargo plane. I’m standing at the head of the plane, near the cockpit door. The plane is travelling at 500 miles per hour. The rear cargo door is open. I point a gun down the length of the plane, towards the rear cargo door opening. I fire a bullet. The bullet travels at 500 mile per hour. What would happen to the bullet as it exits the cargo door, in relation to speed and altitude?
Also, what would happen if I was at the rear of the plane, and fired the same gun towards a an open window at the front?
Do you want to ignore air resistance (because once a bullet exits a moving plane, the air resistance is significant)?
What do you expect to happen? Your bullet is traveling about 500MPH relative to the ground before you fire it. Shortly after, it is traveling at muzzle velocity PLUS the plane’s speed if you fire it in the same direction as the plane, MINUS if the opposite direction.
Relative to the passengers, it’s only muzzle velocity, and the bullet is standing still until fired from their point of view.
In re-reading your question, are you asking what happens as the bullet exits the plane? Let’s ignore air resistance again, but the answer is…in a forward or rearward direction, nothing. Its speed was set when you fired it.
In a downward direction, it began to fall toward earth the minute it left the muzzle and whether it is in or out of the plane, it will continue to accelerate in that direction until it hits something or some other force comes into play.
However, at those air speeds, air resistance will loom large.
…so ignoring wind resistance the bullet flies out the back of the plane then simply drops like a stone when it leaves the cargo door? I’m thinking it does.
Yes, I’m asking what happens to the bullet as it exits the plane.
BACI, thats what I thought. 500mph - 500mph = 0mph. Would that really happen, and at what point in time? If you think about it, it doesn’t seem to make sense that the bullet would pass the cargo door and then suddenly stop and fall immediately to the ground?
If you were watching from the ground, the bullet would have looked like it was standing still the whole time, while the plane is moving at 500 mph. You would see the exit door move away from the bullet, while the bullet is just slowly dropping downward.
(It would also look funny coming out of the gun. Because the gun would look like it was flying away from the stationary bullet.)
Ignoring air resistance, a bullet drops like a stone anyway. But it’s normally traveling so fast that it gets to the target before it has time to drop very far.
Air resistance shortens the trajectory but doesn’t change the basic picture. In other words a bullet does not have any aerodynamic lift, just drag.
Right. More specifically, from the ground’s perspective, the bullet was travelling 500 MPH while it was housed within the gun; same as the passengers and everything else in the plane. When fired, the bullet came apart from that, and as the gun, plane, etc., continued to move horizontally, the bullet stayed put; however, it was no longer prevented from falling by being housed in the gun, and thus begins to fall normally. The plane eventually clears the bullet, and the bullet falls to the Earth at the same horizontal location where it was fired.
You are mixing up frames of reference.
A bullet fired at 500 mph in an airplane going 500 mph (forward) is exactly the same as a bullet dropped from a helicopter which is still in the air.
Air included, the bullet would be moving when it left the airplane. The air resistance in the airplane would slow the bullet down slightly, or from the ground’s perspective, the air in the plane moving at 500 MPH would blow the bullet for a brief moment. So it may be moving at 10 or 20 MPH relative to the ground.
That’s not what happens. From the perspective of the plane, the bullet is going 500mph backwards; however, from the perspective of the ground, the bullet is stationary (in the horizontal direction). If someone on the ground could see the bullet from the moment it left the gun, he would say that it was stationary the whole time and it was the plane that was moving 500mph. You seem to be mixing and matching the frames of reference, which leads to the inconsistency you note.
Imagine a bullet moving at 500mph. Just in front of you, it hits a kevlar barrier, carefully padded behind in order to just absorb all the bullet’s energy. Naturally, the bullet hits the barrier then drops straight to the ground.
Now imagine a bullet moving at 500mph. Just in front of you, it meets a carefully calibrated nozzle that fires a burst of high speed gas at it, decelerating it by exactly 500mph. Naturally, the bullet’s speed is exhausted and it drops to the ground straight down in front of you.
This is exactly what happens with your hypothetical. As you aim the gun, the bullet is moving along (in the gun, in the plane) at 500mph. The gun then acts like the carefully calibrated nozzle mentioned above, firing hot gas at the bullet in an amount exactly enough to slow it to zero. The bullet now falls straight down.
The only difference is that in the plane example, because the plane continues at 500mph, it has moved on and out of the way before the bullet has time to hit the floor of the plane.
So wait, if we change the analogy to a very fast moving ground vehicle, would that mean that it would not be possible for a gunner located at the back of the vehicle to fire a bullet and kill someone on the ground as it moves by? In order for the gun to be effective it would HAVE to fire the bullet at a higher velocity, right?
Assume a slow projectile like a catapult or slingshot, with projectile speed equal to vehicle speed. If you fire backwards at a stationary target , you could hit a pothole in the road by firing as you went over it and the projectile would just fall to the pothole.
If you are firing at something by the side then it makes more sense to fire forward, before you pass the target.
If you fire sideways, perpendicular to the direction of motion, the missile will move in a trajectory 45º to the side of the road having forward motion acquired from the vehicle and lateral motion acquired from the firing.
These are simple vector problems which navigators have to resolve all the time.
If the true wind is 20 Kn from 096º and the boat is doing 6 kn with true course 006º then the apparent wind on the boat is different from the true wind and needs to be calculated. Same thing for currents, etc. Simple vector problems.
