Can anyone tell me the terminal velocity of an average pool ball? (earth gravity, with no air resistance)
And how many feet would it take to reach it?
And, is that velocity faster or slower than the average break speed of an 8 ball player? ( or 9ball)
Your question does not really make sense.
Terminal velocity is usually defined as the velocity in air, how fast is the fastest it will continue to fall.
The velocity of an object falling toward earth from "very far" with no air, assuming they both start with zero relative velocity, and ignoring effects of orbit and other planets, is about 11.2km/s. That is also the velocity (ignoring air) that you need going up to not come back down.
In theory (ignoring air) the distance to reach escape velocity falling is infinite, in practice you get pretty close within a few thousand, or hundreds of thousands of miles, depending on your definition of "close".
Now terminal velocity as normally defined (falling in air, how fast and how long to reach a steady speed) is a more interesting question, and I no longer remember enough math to figure it out. And of course it might vary by whether it was clean, or had some chalk left on it.
But I think it's fair to say that the zero-air speed is a bit faster than anyone can break. Most regular, fairly dense objects' terminal velocity tends to be in the 80-200 mph range (one estimate I saw for a baseball is 95mph), so I would think a pool ball might be comparable.
I found a very old article saying a break shot is under 30mph, so I think they are slower than air-slowed terminal velocity as well.
Article here. I can't vouch for its accuracy, google gave it to me.