Cars on parked cliffs?

Question

A car is parked on a cli® overlooking the ocean
on an incline that makes an angle of 22:0±
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 4.60 m/s^2 and travels 48.0 m to the edge of
the cliff. The cliff is 38.0 m above the ocean.
The acceleration of gravity is 9.81 m/s^2.
a) How long is the car in the air? Answer
in units of s.

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

Answer 1

I'll help but I'm not going to give you the complete answer. You can figure it out.

First, given the acceleration on the slope, 4.6 m/s^2, and the distance, 48 meters, you can use vf^2 = v0^2 + 2*4.6*48
or vf = 0 + sqrt(2*4.6*48)
Using trigonometry, you can compute the vertical component of this as sin(22 degrees)*vf

This is now used as the initial downward velocity, and you can again use the formula
vf(ocean)^2 = vf^2(slope answer) + 2*9.81*38
Once you have vf(ocean) then you can use
time = 2*38m/(vf(slope)+vf(ocean) to get the number of seconds the car is in the air.

For the car position relative to the base of the cliff, assume a constant horizontal speed of cos(22 degrees)*sqrt(2*4.6*48), times the number of seconds in the air, to compute the distance.