A bunch of fake Indians are on a ship docked in Boston Harbor and want to push a

Question

A bunch of fake Indians are on a ship docked in Boston Harbor and want to push all the crates of tea over the edge of the ship, into the water. They set up a ramp to do it. The ramp is 10.0 m long, and its end is propped up so that it is 5.00 m above the deck. The deck of the ship is 15.0 m above the surface of the water. Each crate of tea has a mass of 50.0 kg. They push the crates of tea all the way up the 10.0 m ramp, two at a time, starting from the bottom. (Ignore any friction.)
a. If the Indians push the crates up the ramp with an acceleration of 11.0 m/s2, what is the horizontal distance from the edge of the plank, d, that the crates hit the water?
b. On one of their runs up the ramp, the top crate falls off half way up the ramp. Without pause, the Indians continue to push the remaining crate up the ramp. But with only one crate to push, the acceleration now instantaneously doubles. What is the horizontal distance from the edge of the plank, d, that the single remaining crate now hits the water?

Explanation / Answer

speed of crates at the top of ramp

V^2 = u^2 + 2aS

V^2 = 0 + 2*11*10

V = 14.83 m/s

net height above the sea level(h) = 5+15 = 20 m

horizontal distance from the edge of the plank, d, that the crates hit the water

d = V sqrt(2h/g)

d = 14.83 sqrt(2*20/9.81)

d = 29.94 m

b.)

speed of the ramp half off the way

V^2 = U^2 + 2aS

V^2 = 0 + 2*11*5

V = 10.48 m/s

now,

Velocity at top of ramp

V'^2 = V^2 + 2aS

V'^2 = 10.48^2 + 2*22*5

V' = 18.16 m/s

d = V' sqrt(2h/g)

d = 18.16 sqrt(2*20/9.81)

d = 36.67 m

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