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 60.0 m to the edge of the cliff, which is 50.0 m above the ocean.

Required:
Find the car’s position relative to the base of the cliff when the car lands in the ocean.

Answer 1

The position is
`P = 47.4 \ m` relative to the base of the ocean

Step-by-step explanation:

From the question we are told that

The angle made by the incline with the horizontal is
`\theta = 24.0 ^o`

The constant acceleration is
`a = 3.82 \ m/s^2`

The distance covered is
`d = 60.0 \ m`

The height of the cliff is
`h = 50 .0 \ m`

The velocity of the car is mathematically represented as

`v^2 = u^2 + 2ad`

The initial velocity of the car is u= 0

So

`v^2 = 2ad`

substituting values

`v^2 = 2 * 3.82 * 60`

`v = 21.4 \ m/s`

The vertical component of this velocity is

`v_v = -v * sin(\theta )`

substituting values

`v_v = -21.4 * sin(24.0)`

`v_v = -8.7 \ m/s`

The negative sign is because is moving in the negative direction of the y-axis

The horizontal component of this velocity is

`v_h = v * cos (\theta)`

`v_h = 21.4 * cos (24.0)`

`v_h = 19.5 \ m/s`

Now according to equation of motion we have

`h = v_v*t - (1)/(2) * g t^2`

substituting values

`50 = -8.7 t - (1)/(2) * 9.8 t^2`

`4.9t^2 +8.7t -50 = 0`

using quadratic equation we have that

`t_1 = 2.42\ s \ and\ t_2 = -4.20\ s`

given that time cannot be negative

`t = 2.42 \ s`

The car’s position relative to the base of the cliff when the car lands in the ocean is mathematically evaluate as

`P = v_h * t`

substituting values

`P = 19.5 * 2.43`

`P = 47.4 \ m`