Lord Beckett and members of the EIT Co. spot the Black Pearl in the distance making its way towards land. As a precaution, Lord Beckett orders a cannon to be shot towards Captain Jack Sparrow. The cannonball goes a horizontal

distance of 275 m after leaving the 100. m high cliff. If the cannonball was launched from horizontal, at what velocity

did it leave the cannon? At what velocity (resultant speed + angle) did it hit the sea? (Clearly, they missed the ship!)

Show work

Explanation:

The cannonball goes a horizontal  distance of 275 m . It travels a vertical distance of 100 m

Time taken to cover vertical distance =  t ,

Initial velocity u = 0

distance s = 100 m

acceleration a = 9.8 m /s²

s = ut + 1/2 g t²

100 = .5 x 9.8 x t²

t = 4.51 s

During this time it travels horizontally also uniformly so

horizontal velocity Vx = horizontal displacement / time

= 275 / 4.51 = 60.97 m /s

Vertical velocity Vy

Vy = u + gt

= 0 + 9.8 x 4.51

= 44.2 m /s

Resultant velocity

V = √ ( 44.2² + 60.97² )

= √ ( 1953.64 + 3717.34 )

= 75.3 m /s

Angle with horizontal Ф

TanФ = Vy / Vx

= 44.2 / 60.97

= .725

Ф = 36⁰ .

answer: 0.20

explanation: this can only be the answew because all the other measuremnets are not close