Wow.... that must be a long barrel on the cannon.
4kg 60N 50degrees 1s
60 = 4a a = 15ms
-2 15 * 1 = 1 v = 15ms
-1 That means that the barrel is 7.5 meters long, assuming force is constant along it and has an instant dropoff at the end. Anyway, now in the vertical direction, the velocity is sin(50) * 15 and in the horizontal, it's cos(50) * 15, because triangles are triangles. What we care about for now is the vertical velocity (11.491m/s). I'm assuming that the bottom of the barrel is touching the ground, so the top is at sin(50) * 7.5 meters (5.745m).
With these, I know that the peak of the flight will be reached at 11.491/9.8 seconds (1.173s) and will be 5.745*1.173 meters above the end of the barrel (12.484m above the ground).
Because v
2 = u
2 + 2ax, the velocity at the ground level is equal to the square root of 2ax (u = 0) and 2ax = 244.686, so v = 15.642m/s.
Time to ground = time to accelerate to 15.642m/s = 15.642 / 9.8 = 1.596s
The total time in the air is 1.173 + 1.596 seconds (2.769 seconds).
Now, to figure out the distance travelled, we can simply say that the horizontal velocity (9.642m/s) times the time (2.769s) is the distance. This gives us a distance of 26.699m.
Am I right?????? I hope I didn't screw up (I know I could have saved some equations if I looked up formulas, but you know... fun!)