How do I estimate the transmission distance of a 50 watt 2m radio?

Providing that the ground between transmitter and receiver is flat the following formula gives the distance to the horizon:

\$ D =1.2\sqrt{H} \$

where H is the antenna height above the ground in feet and D is the distance to horizon in miles. You then have to double this distance so that both antennas see the horizon.

So with antenna height of 7 meters (approximately 22 feet) this gives

\$ 1.2 \sqrt{22} = 1.2 * 4.7 = 5.6\mbox{ miles} \$

Double this to give 11.2 miles.

This assumes that both stations are at the same height above sea level. If one station is higher than the other, the answer will be greater than 11.2 miles. Providing that there are no obstacles between the two stations (buildings, hills, mountains, etc.) it should be possible to cover this distance with only 1W.

If you need it to be reliable, then you are talking about line of sight, if not direct than to a repeater on a tall building, mountaintop, or satellite. In most of these cases, 50 W should be sufficient with non-terrible antennas and no undue interference.

There are some occasional over-the-horizon atmospheric propagation modes for 2 m (and even meteor ionization trails for infrequent packets) which you can read about in the ARRL handbook, but you can't depend on them.

You would need a huge antenna farm to get enough directionality to do moonbounce with 50 watts.