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 16° 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 4.1 m/s^2 and travels 47 m to the edge of the cliff. The cliff is 40 m above the ocean.
a. How long is the car in the air? The acceleration of gravity is 9.81 m/s^2. Answer in units of s.
2. What is the car's position relative to the base of the cliff when the car lands in the ocean? Answer in units of m.

Explanation / Answer

the information allows us to figure out the horizontal and vertical speed of the car when it goes off the edge of the cliff we find that speed from: vf^2=v0^2+2ad where vf is final speed v0=initial speed = 0 a = accel = 3.82 m/s/s d=distance = 35m so the speed on leaving the cliff is: vf^2=0+2(3.82)(35) vf^2 = 267.4 vf=16.35 m/s now, we need to find the components of the car's velocity as it leaves the cliff, they are: v(horizontal) = 16.35 cos 24 = 14.9m/s v(vertical) = -16.35 sin 24 = -6.65 m/s we need to find the time the car is in the air, for this we use the equation of motion: y(t)=y0+v0y t - 1/2 gt^2 y(t)=height at any time t, y==initial position v0y=initial y speed so we have: y(t)=20-6.65t-4.9t^2 we want to find how long it takes for the car to reach y=0: 0=20-6.65t-4.9t^2 this is a quadratic equation with solution t= 1.45s since the horizontal speed will not change once the car leaves the clilff (since there are no horizontal forces acting), we have that the horizontal distance traveled in 1.45 s is: x=14.9m/s x 1.45s =21.6m

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