A car is parked on a cliff overlooking the ocean on an incline that makes an ang

Question

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0° below the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline with a constant acceleration of 3.82 m/s2 for a distance of 30.0 m to the edge of the cliff, which is 50.0 m above the ocean. (a) Find the car's position relative to the base of the cliff when the car lands in the ocean. (b) Find the length of time the car is in the air.

Explanation / Answer

First, find the final velocity of the car once it reaches the base of the cliff. v^2=2ax = 2*3.82*30.0 = 229.2 v=15.139 m/s v_x=15.139cos24.0 = 13.8 m/s This velocity is the velocity of the car in the x direction once it falls. Next, we find how long it will take for the car to it the water. Acceleration due to gravity is 9.81 m/s^2. If the car is going 15.139 m/s at an angle of 24.0 deg below horizontal, then v_iy=15.139sin24.0 = 6.158 m/s y=v_iy+1/2*at^2 ==> 30.0=6.158+0.5*(9.81)t^2 ==> t = sqrt((30.0-6.158)/(0.5*9.81)) = 2.205 s Now, just multiply v_x by t: 13.8*2.205 = 30.4 m (answer for part a) We already calculated t = 2.205 (answer for part b)

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