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 14.0 below the horizontal. The brakes fail, and the car rolls from rest down the incline for a distance of 33.9 m to the edge of the cliffm which is 54.1 m above the ocean, and, unfortunately,d dcontinues over the edge and lands in the docean.
Part 1 out of 2
a) 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 = 14 o Height of the cliff h = 54.1 m / s Distance moved S = 33.9 m Accleration a = g sin                      = 2.37 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                    = 12.67 m / s In falling : --------- Initial velocity in vertical direction U = v sin                                                        = 3.067 m / s Accleration a ' = g                        = 9.8 m / s 2 From the relation h = Ut + ( 1/ 2) gt 2                    54.1 = 3.067 t + 4.9 t 2 4.9 t 2 + 3.067 t -54.1 = 0 t = {-3.067 ± [ 3.067 2 -(4x 4.9 x -54.1 ) ] } /2(4.9) = { -3.067 ± 32.7 } / 9.8   = 3.024s So, required distance = Horizontal velocity x time                                  = v cos x t                                  = 37.18 m Initial velocity u = 0 Angle = 14 o Height of the cliff h = 54.1 m / s Distance moved S = 33.9 m Accleration a = g sin                      = 2.37 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                    = 12.67 m / s In falling : --------- Initial velocity in vertical direction U = v sin                                                        = 3.067 m / s Accleration a ' = g                        = 9.8 m / s 2 From the relation h = Ut + ( 1/ 2) gt 2                    54.1 = 3.067 t + 4.9 t 2 4.9 t 2 + 3.067 t -54.1 = 0 t = {-3.067 ± [ 3.067 2 -(4x 4.9 x -54.1 ) ] } /2(4.9) = { -3.067 ± 32.7 } / 9.8   = 3.024s So, required distance = Horizontal velocity x time                                  = v cos x t                                  = 37.18 m

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