Almost any elementary physics text would have a discussion of this problem, assuming that the earth is a plane and ignoring atmospheric effects.

The assumption that the earth is a plane introduces no significant error, but atmospheric effects would be noticeable.

I think the following would be valid for a plane surface and no atmosphere.
  • HorizontalDistance = Velocity*Time*cos(Angle)
  • VerticalDistance = Velocity*Time*sin(Angle) - GravityAcceleration*Time2/2
Distances measured from end of the cylindrical hole in the cannon barrel. Correct units must be used.

I think the GravityAcceleration is 32.16 feet/seconds2

In metric it might be about 980 cm/seconds2

I would not advise using any of the above information without verifying from some reliable source.

I do not know how to calculate the effects due to the atmosphere. For golf balls, the direction of the wind is very significant, but I am not sure that direction is significant for artillery projectiles. I am pretty sure that density and composition of the atmosphere of the air is significant. I think the amount of water vapor in the air would have an effect.