If I dropped a bullet from a stationary platform in the air, it will accelerate to terminal velocity (lets say 120mph). If I shot a bullet downward (assuming faster than terminal velocity) it will decelerate to terminal velocity.
Is there a speed that it can be shot that the velocity will not change? If I shot it at 120mph, would it maintain that speed…or would it slow down due to air friction then speed up again…or would it speed up due to gravity acceleration added in then slow down?
By the way, did you ever find it odd that water a few feet from coming out of a horizontal hose seems to have more force than when it comes immediately out of the hose? Try it, as the water falls away it seems to pick up force (speed?)…I’d rather get hit in the face with water right out of a hose than be a few feet away as its arcing down.
It’s not arcing down because it’s losing the force from the hose (well, much, anyway – air resistance does slow it down some), it’s arcing down because it’s adding the force from gravity. So yes, it’s total momentum is higher once it starts to drop.
the projectile is stable in flight (not tumbling, etc),
terminal velocity is 120mph
the air it’s traveling is uniform
the gravitational field is uniform
No other forces act on the projectile once it leaves the barrel of the gun
then yes, if fired at 120mph it will remain exactly at that speed with no further accel/decel. At that point the force of gravity is exactly balanced with the drag force and it’s in equilibrium until acted on by an outside force.
Terminal velocity is plenty higher at high altitudes than at ground level, due to lower air pressure. So an object with ground level terminal velocity 120 mph would, if shot straight down at that speed from, say, an altitude of 10 km initially pick up some speed before slowing down again.
Terminal velocity is different depending on the shape and mass of the object, the density of the air, and the movement of the air relative to the object (i.e., if the frame of reference for speed is relative to the ground).
It looks like the terminal velocity for a bullet is actually 204 mph through near-surface air density.
If you could shoot the bullet at 204 mph, then in a simple theoretical model it would just continue on at exactly 204 mph. In the real world, the transition from the inside of a barrel to the outer world would probably produce some ripple effect in the air that might cause the air to move with and/or against the direction that the bullet is traveling. So, if measured at a fine enough level, the speed of the bullet will almost certainly fluctuate up and down until it has gotten out of range of the air effects of the muzzle blast. After that, wind effects would take over, as would any toppling of the bullet, slightly shifting the speed and terminal velocity of the bullet subtly.
The bullet would probably always be very close to 204 mph, but measured finely enough, it would fluctuate in a fairly chaotic fashion.
ETA: Gravity factors in as well. It’s not exactly even across the surface of the planet. So terminal velocity might be different than average based on that.
This was amply demonstrated by Felix Baumgartner when he skydived from 120,000+ feet. He reached supersonic speeds before entering thicker air in the lower part of the atmostphere, where the increased drag decelerated him to less than 200 MPH before he deployed his parachute.
The problem is complicated by the fact that the terminal velocity of a dropped bullet will be much less than the terminal velocity of a bullet fired point down from a gun due to the bullet’s spin. A dropped bullet will tumble, a fired bullet won’t.
I think that might be a large part of it. The further away you get, the water stream breaks up into large ‘pellets’, so you are pummeled by pellets with nothing in between. The average force could be the same as for an unbroken stream, but the fact that its not a single nice easy push, but a ‘splat’ in the face intermingled by nothing, makes it more traumatic.
For this same reason, the water’s drag force against an object near the hose outlet could actually be less than further away. IOW it might not just feel like less force, it could actually be less force.
The principle can be clearly seen in laminar flow faucets. Due to the smooth flow, the stream tends to follow the contours of an object. IOW the total drag force on the object is less than with turbulent flow at the same rate and velocity. With a water hose the same laminar-to-turbulent transition could be happening naturally, so the force could actually increase (not just feel like an increase) further away from the nozzle: http://homeenergy.org/images/picts/98073601.gif
