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 15.0° 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 2.43 m/s2 for a distance of 28.0 m to the edge of the cliff. The cliff is 68.0 m above the ocean.

With what speed does the car leave the cliff?
m/s
Find the car's position relative to the base of the cliff when the car lands in the ocean, and
m
How long is the car in the air? (Hint: If you would like to avoid a quadratic equation, solve the next part first.)
s
With what speed does the car hit the water?
m/s

Explanation / Answer

You can use this equation to find Vf as the car leaves the cliff and begins to fall:

2ad = Vf ^ 2 - Vi ^2

2*2.43*28 = Vf ^ 2 - 0

Vf = 11.67 m/s

Now you can split this velocity up into its components.

As the car leaves the cliff, it forms a triangle of velocities with 11.67 on the hypotenuse, and

Vx = 11.67*cos(15), Vy = 11.67*sin(15)

These are the initial velocities in the horizontal and vertical directions, respectively.

Use d = Vi*t + 0.5at^2 in the Vertical to solve for time.

- 68 = - 11.67*sin (15) * t + 0.5(-9.8)t^2

- 68 = - 3.02 * t + - 4.9*t^2

4.9*t^2 + 3.02 * t - 68 = 0

Solving, we get

t = 3.43 s

You can then use this time in D = VT in the horizontal,

to solve for D.

D = VT
D = 11.67*cos(15) * 3.43
D = 38.66 m

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