How can I calculate where a ball will travel/land if it projected out of a cannon?

For a 45 degree inclined angle, the vertical speed is half the ball speed. So that is 0.5 m/sec.

Use the gravitational acceleration formula to determine how much time it would take a ball to return to earth if it's thrown vertically at 0.5 m/sec.

Use that answer to calculate how far a ball travels with a forward velocity of 0.5 m/sec (which is the horizontal speed of the ball).

If we assume no drag forces, the curve of a trajectory will form a parabola open downward.

To map that curve you need the x(t) = Ux t position along the X axis, and the y(t) = y(0) + Uy t - 1/2 g t^2 along the Y axis. Each pair p(x,y) is a point on the curve. t is some given time after the shot when t = 0 in seconds usually.

Ux = U cos(theta) is the horizontal speed along the X axis. Uy = U sin(theta) is the vertical speed along the Y axis. U = 1 mps is the muzzle speed and theta = 45 deg is the angle re the horizontal of the shot. g is g, usually 9.81 m/s^2 or 32.2 ft/s^2 depending on the units you're using.

You now have everything needed to graph the curve on Cartesian Coordinates. I'll get you started:

t______ X______ Y

0.000 0.000 2.000 <Note I assumed a muzzle height y(0) = 2 meters

0.100 0.071 2.022

0.200 0.141 1.945

0.300 0.212 1.771

Also note, at U = 1 mps muzzle speed, that cannon ball isn't going very far. It almost immediately starts to drop right out of the muzzle. I used g = 9.81 m/s^2 for this example. Just keep adding tenth of a second until Y = 0, which is impact. X and Y are in meters.

The vertical component vy = v * sin(theta) determines how long it is in the air, using the equations of accelerated motion.

For instance, you can use vf = vi + at => 0 = vy + at with a = -9.8 m/sec^2 to find out the time for the vertical velocity to be 0. That's the time to reach the maximum height. Then double it.

Once you know how long it's in the air, use d = vx * t with vx = horizontal component = v * cos(theta) to find how far it goes.