What is the formula for the distance a projectile will travel with a constant initial velocity?

The formula for the distance a projectile will travel fired with initial velocity v and angle ? above horizontal is

d = v²/g * sin2?

the maximum distance is obtained at ? = 45º

g = 9.8 m/s²;  v in m/s and d in m.

The unambiguous equation for level ground is:

R = [Vi²*sin(2?)]/g

Uneven ground is a bit trickier.  Email me when you come to that part.  I have all the eqs handy.

You could set the intial coordinates of the projectile to be at (0,0) and then you can find the horizontal range by using the recipe as follows:

In the horizontal direction the accelertion is zero since the cannon is not dependent on any acceleration in the vertical direction.
dx = v0*cos?*t + 0 + 0

In the vertical direction you do have acceleration so
dy = v0*sin?*t - (1/2)gt^2

When the height of the distance as given by above goes to zero (when it falls to the surface) then you get (1/2)gt^2 = v0*sin?*t and then basically you have the time, so substitute that back into dx and it becomes v0^2 *sin(2?)/g

The formula already given is okay for short range artillery where air resistance, coreolis effect and curvature of the Earth are insignificant. If you want to hit a small target a 1000 km away, you'll need more complex formulas.