your friend\'s car is parked on a cliff overlooking the ocean on an incline (hat

Question

your friend's car is parked on a cliff overlooking the ocean on an incline (hat makes an angle of I8.1degree below the horizontal. The brakes fail, and the car rolls from rest down the incline for a distance of 30.7 m to the edge of the cliff, which is 55.9 m above the ocean, and, unfortunately, continues over the edge and lands in the ocean. Part 1 out of 2 Find the car's position relative to the base of the cliff when the car lands in the ocean. The car falls meters from the base of the cliff.

Explanation / Answer

Initial velocity u = 0 Angle = 18.1 o Height of the cliff h = 55.9 m / s Distance moved S = 30.7 m Accleration a = g sin                      = 3.044 m / s 2 Velocity of the car at the edge of the cliff v = ? From the relation v 2 - u 2 = 2aS From this v = [ 2aS ]    Since u = 0                    = 13.67 m / s In falling : --------- Initial velocity in vertical direction U = v sin                                                        = 4.246 m / s Accleration a ' = g                        = 9.8 m / s 2 From the relation h = Ut + ( 1/ 2) gt 2                    55.9 = 4.246 t + 4.9 t 2 4.9 t 2 + 4.246 t -55.9 = 0 t = {-4.246 ± [ 4.246 2 -(4x 4.9 x -55.9 ) ] } /2(4.9) = { -4.246 ± 33.37 } / 9.8   = 2.972 s So, required distance = Horizontal velocity x time                                  = v cos x t                                  = 38.61 m

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