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 17.0° 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 2.75 m/s2 for a distance of 65.0 m to the edge of the cliff. The cliff is 45.0 m above the ocean.

Find the car's position relative to the base of the cliff when the car lands in the ocean.

Find the length of time the car is in the air.

Explanation / Answer

lets use A as the point of start, B as the point of the edge of cliff and C as the final position

From the A-B interval, it travels from rest to edge of cliff over a dist of 65m (along the cliff)

The component that is causing its weight

So the resolving the acceleration is x-y direction:

x-direction ( along the clifft) => a = g sin 17 = 2.86 m/s^2

Using v^2 = u^2 + 2 as (starts from rest u = 0)

v^2 = 2*2.86* 65 = 371.8

v = 19.28 m/s (along the cliff)

Now moving on the B-C interval where we need to apply projectile motion

Before that we have to resolve again for velocity

v_x = 19.288 cos17 = 18.43 m/s

v_y = 10.08 sin 17 = 5.63 m/s

We solve the y direction first to determine the time when it lands

s = ut + 1/2 gt^2

45 = 5.63 t + 1/2(9.81) t^2

4.9t^2 + 5.63 t -45 = 0

t = 3.58 secs --------<<<<<<<<<<<<<<<<<<<<Answer toi part B

Now solving for the x-direction:

s = u t (there is no acceleration in x -direction)

s = 18.43 * 3.58 = 66.06 m ----------------<<<<<<<<<<<<Answer to part A

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