[mathking]

I wouldn't bother with computations of drag. At 12 m/s with a circular ball flung by some mechanical system, the influence of drag is probably not going to be distinguishable from other sources of variation.

Why not just do some basic calculations with pencil and paper and then make a spreadsheet. All you really need is basic physics:
v = initial_velocity + at d = vt + .5*a*t^2 quadratic formula.
(OK, you need sin and cos of your angle too.)

A perhaps more interesting question than what is the farthest a ball can be thrown is what is the minimum angle of elevation required to get the ball into the center goal at all?

Example:

Assume an initial "muzzle velocity" of 12 m/s and an angle of 20 degrees:
Upward velocity = 12 * sin(20) = 4.104 m/s
So what is the maximum height above the gun this ball will reach?
Well it will reach this max height when v = 0, so:

0=4.104 - 9.8t
9.8t = 4.104
t = 4.104/9.8 = 0.419 seconds

d = 4.104*0.419 + (.5)(-9.8)(0.419^2)
d = 1.719 - .860
d = 0.859 m = 33.8 inches

When you release the ball, remember that no part of the robot can be above 60 inches. So the bottom of the ball at the release point may be well below 60 inches. But let's assume the top boundary condition of a release at 60 inches.

60 + 33.8 = 93.8 inches = 7 ft 9.8 inches = well below the goal.