What latitude is needed before you can reliably see the globular cluster Omega Centauri, say it reaches 20 degrees above the horizon? What about if you are up on a hill looking down, what's the theoretical highest latitude where you could have any chance of seeing it?
Since Omega Centarui is at a declination of -47.5 degrees, You would need to be at a lattitude of 23.5 degrees North (i.e. on the Tropic of Cancer) for it to get to 20 degrees above the horizon when it crosses the meridian (due south).
In theory, if you had a perfectly flat horizon (and not atmosphere), you could see it just on the horizon as it crosses the meridian at a latitude of 42.5 degrees north.
To see it from a hill with a negative horizion the lattitude would depend on the amount of negative horizon you had.
This assumes that there is no atmosphere. Even at 30 degrees above the horizion you are looking through twice as much air as straight up. At 20 degrees you are looking through 3 times as much air and it gets worse as you go to the horizion, this will effectively limit the minimum angle you can look at. Omega Cen is a 3.7 magnitude object so it's fairly bright but even so, you'll loose it in the atmosphere before you get all the way to the horizon.
Omega Centauri has been observed by a group of my friends in the London Centre of the Royal Astronomical Society of Canada, from atop Eagle Bluff on the north shore of Lake Erie, around 42° 30' N latitude. This was about 7 or 8 years ago, and the cluster just cleared the southern horizon.