However per the original from firefighter - I can think of no way that ''g'' can accelerate a downward travelling projectile to achieve anything like original MV - which we might assume to be from a carbine 9mm - something like 1200 and more fps
Hrm
I hadn't thought about it that way, though of course I should have. You're right: once the bullet has reached its maximum altitude, the only things we have to worry about are wind resistance and g. So, using 1200 fps (both because I trust your estimate on speed, and because that's not much past Mach 1, and I know wind resistance dynamics change drastically at Mach 1), for the round to accelerate to that speed due to g, even ignoring drag, it would have to be falling for 37.5 seconds (unless my math is off, of course). This seems like a really long time, to me.
By the same token, if it started out at 1200 fps, it would take 37.5 seconds for gravity to bring it to a halt, too. It won't take that long in atmosphere, of course, because drag will add to gravity's 9.8N (with apologies for mixing SI and English units). On the way back down, then, it won't have to fall as far, which also means it doesn't have as long to accelerate, so you must be right - it can't possibly land with the same velocity it had going up.
And, in a Eureka moment, I just understood why. On the way up, you've got three forces: the initial acceleration (up), wind resistance (down), and gravity (down). On the way down, you've got two forces: wind resistance (up), and gravity (down). Given that gravity's a constant, and assuming that the drag curves roughly match each other, the "down" forces on the way up are higher than the "down" force on the way down.
Rusty, my Aunt Petunia's patootie...you just answered for me a question that's been mildly bothering me for years.