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.