Cannons and golf balls - Question:

I have a neighbour who has an old fashioned cannon. On holidays, he points the barrel so that it is almost perpendicular to the ground, and he fires a golf ball into the air.
It takes 15 seconds from the time the golf ball leaves the barrel until we see the ball hit the ground.
What formula do I use in order to determine how high the golf ball went? I know that acceleration (9.8 meters per second/per second) has something to do with it, but that's all I remember from high school physics. Maybe the formula is as simple as determining how far the golf ball would fall in 7.5 seconds? That's assuming that the barrel is truly perpendicular (no angle)

However, if you've got a formula that would take the angle into account, I'd love that, too. I'm assuming we could help figure out the angle by determining how far away the golf ball landed from the cannon (we'll cheat and assume it's a calm, windless day)

There are a number of useful formulae but all of them require you to know something about the muzzle velocity of the cannon. (All of them ignore the air resistance which with a spherical ball will make the problem much more difficult – it is usually reckoned that the drag due to the air is proportional to the square of the speed of the ball at any time).
Anyway to the other formulae.

Height and range of a projectile.

1. Fired vertically upwards
(a) Knowing the time that it takes for the ball to reach the ground again (2t)
Speed at the top of the path (v) is zero (at the top of the path it is not moving upwards).
Height risen (h) = ½ gt2 where t is the time of flight to the top.
As you say g is the acceleration due to gravity (9.8 m/s2)
So if the ball, takes 15 seconds to return to the ground (2t = 6 so t = 7.5 s)
Therefore height risen (h) = ½ x 9.8 x 7.52 = 275.6 m. Impressive!
You can also work out the muzzle velocity (u) of the cannon.
This time the formula is: v = u + gt with the letters meaning the same as given above
0 = u + (-9.8)x7.5 u = 73.5 m/s
Notice the negative sign in front of the g, this means that the ball is decelerating as it goes upwards as is the case.
So, ignoring air resistance and just measuring the time of flight (3s in the example) we now know the height reached (275.6 m) and the muzzle velocity (73.5 m/s).

2. Now to the angled flight (again assuming no air resistance).
This time the formulae are a bit more complicated with the angle of the cannon to the horizontal appearing as A.
Range (R) = u2sin 2A/g
Maximum height reached (h) =[u2sin2 A]/2g
Again an example but this time we will assume the muzzle velocity of 29.4 m/s from the first example and an angle of 40o between the cannon barrel and the ground.
Range (R) = [73.52 x sin (80)]/9.8 = 542.9 m - over half a kilometre!
Maximum height reached (h) =[73.52 x sin2 40]/2x9.8 = 113.9 m
A very impressive cannon.
The difference between the predicted and actual values will be due to air resistance and wind.

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