Question

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 23.1 ◦ 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 2.52 m/s 2 and travels 37.5 m to the edge of the cliff. The cliff is 21.1 m above the ocean. Find the car’s position relative to the base of the cliff when the car lands in the ocean. The acceleration due to gravity is 9.8 m/s 2 . Answer in units of m.

Answer 1

Answer:20.17 m

Step-by-step explanation:

Given

Angle of inclination
`=23.1 ^(\circ)`

acceleration of car
`=2.52 m/s^2`

Distance travel 37.5 m to the edge of the cliff

Velocity after travelling 37.5 m

`v^2-u^2=2as`

`v^2-0=2* 2.52* 37.5`

v=13.74 m/s

Now the car is launched at an angle of
`23.1^(\circ)` with the horizontal

Vertical distance traveled

`21.1=vsin23.1t+(gt^2)/(2)`

`4.905t^2+5.39t-21.1=0`

t=1.596 s

Thus car position relative to the base of the cliff
`=vcos23.1* t`

`=13.74* cos23.1* 1.596=20.17 m`