Assuming the dread pirate roberts never misses, how far from the end of the cannon is the ship that you are trying to hit (neglect dimensions of cannon) ? answer in units of m.

First let's find the acceleration required in the barrel to speed the ball up from 0 to 83 m/s in a distance of 2.17 m. We know the force the cannon exerts on the cannonball is 20000 N; if we can find this acceleration then we can use F = ma to find the mass. We can find the acceleration using one of the kinematic equations of motion. We have: u = initial speed = 0 m/s v = final speed = v0 = 83 m/s d = distance = 2.17 m a = acceleration = ? v² = u² + 2ad. Since u = 0, this reduces to v² = 2ad and rearranges to a = v²/2d = 83²/2*2.17 = 83²/4.34 = 1587.327 m/s². Now F = ma, so m = F/a = (20000N) / (1587.327 m/s²) = 12.6 kg. For part 2, use the Range Equation: If R is the horizontal distance the cannonball travels, v = v0 = the initial velocity = 83 m/s g = acceleration due to gravity - 9.8 m/s² x the launch angle relative to the horizontal, then R = (v²sin (2x)) / g. So R = (83²sin (2*37)) / 9.8 = (6889sin74) / 9.8 = 676 m. So the target ship is 676 m away.