Question #4904

A cannon mounted on a pirate ship fires a cannonball at 125 m/s horizontally, at a height of 17.5 m above the ocean surface. Ignore air resistance. (a) How much time elapses until it splashes into the water? (b) How far from the ship does it land?
(a)

Expert's answer

A cannon mounted on a pirate ship fires a cannonball at Vh=125 m/s horizontally, at a height of 17.5 m above the ocean surface. Ignore air resistance.

(a) How much time elapses until it splashes into the water?

Initial vertical velocity Vv of a cannonball is equal to zero cause cannon fires horizontally. The gravity is acting on a cannonball, and Vv(t) = gt. So, (Vv^2)/2 = (gt^2)/2 = 17.5. Here we get T = sqrt(2*17.5/g) = sqrt(2*17.5/9.8) = 1,9 s.

(b) How far from the ship does it land?

While falling down the cannonball will cover the distance D = Vh*T = 125*1.9 = 237.5 m.

(a) How much time elapses until it splashes into the water?

Initial vertical velocity Vv of a cannonball is equal to zero cause cannon fires horizontally. The gravity is acting on a cannonball, and Vv(t) = gt. So, (Vv^2)/2 = (gt^2)/2 = 17.5. Here we get T = sqrt(2*17.5/g) = sqrt(2*17.5/9.8) = 1,9 s.

(b) How far from the ship does it land?

While falling down the cannonball will cover the distance D = Vh*T = 125*1.9 = 237.5 m.

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