A automobile is parked on a cliff overlooking the ocean 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 space defective. The automobile rolls from rest down the incline through a continuous acceleration of 4.8 m/s 2 and travels 43 m come the sheet of the cliff. The cliff is 30 m above the ocean. (a)How lengthy is the vehicle in the air? The acceleration of heaviness is 9.81 m/s 2 . Prize 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 basic of the cliff as soon as the car lands in the ocean? answer in units of m.


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At the bottom of the incline v2= u2 + 2as,,, u=0. A = 4.8, s = 43, for this reason v = 20.3 m/s

So the horizontal velocity in ~ that suggest will it is in 20.3 cos 17 = 19.4 m/s, and also the upright velocity bottom 20.3 sin 17 = 5.94 m/s

To uncover the time in this air, this is the moment it take away the vehicle to autumn 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

giving t = 3.2 s indigenous graphing it and also using trace

So horizontal distance took trip = 3.2 x 19.4 = 62.08 m

Mike


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