Your friend\'s car is parked on a cliff overlooking the ocean on an incline that

Question

Your friend's car is parked on a cliff overlooking the ocean on an incline that makes an angle of 18.1degree 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. 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. Find the length of time the car is the in the air. The car is in the air for seconds.

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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