A car is parked on a cliff overlooking the sea on an incline that provides an angle of 17 degrees below the horizontal. The negligent driver leaves the automobile in neutral, and the emergency brakes are defective. The auto rolls from rest down the incline with a constant acceleration of 4.8 m/s 2 and travels 43 m to the edge of the cliff. The cliff is 30 m over the sea. (a)How long is the auto in the air? The acceleration of gravity is 9.81 m/s 2 . Answer in systems of s.

You are watching: A car is parked on a cliff overlooking the ocean on an incline

(b) What is the car’s position relative to the base of the cliff as soon as the auto lands in the ocean? Answer in systems of m.

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At the bottom of the incline v2= u2 + 2as,,, u=0. a = 4.8, s = 43, so v = 20.3 m/s

So the horizontal velocity at that suggest will be 20.3 cos 17 = 19.4 m/s, and also the vertical velocity downward 20.3 sin 17 = 5.94 m/s

To discover the moment in this air, this is the time it takes the car to loss 30 m.

s = ut + 0.5 a t2

30 = 5.94 t + 0.5 x 9.81 t2

4.905 t2 - 5.94t - 30 = 0

offering t = 3.2 s from graphing it and using trace

So horizontal distance travelled = 3.2 x 19.4 = 62.08 m

Mike

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