What's given:
Cliff angle with the horizontal, θ = -19°
Distance on the cliff, d = 50m
Height of cliff, h = -50m
(0 being on top of the cliff)
acceleration before leaving cliff, a = 3.06 m/s²
acceleration due to gravity, g = -9.8 m/s²
Car was stationary at the beginning, so
initial velocity, u = 0
First calculate the speed of the car, v, when it left the cliff using
v²-u² = 2ad
At the point of leaving the cliff,
horizontal velocity, vh = vcos(θ)
vertical velocity, vv = vsin(θ)
(note the sign of θ)
We now can calculate the time in the air, t using
h = vv*t + (1/2)gt²
The horizontal distance from the cliff, S can be calculated by the product of the initial horizontal velocity, vh by the time in the air.
Post if you need explanations or answer check.
This has been posted before but I need someone to elaborate please.
A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 19.0° 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 3.06 m/s2 for a distance of 50.0 m to the edge of the cliff, which is 50.0 m above the ocean. Find the following.
(a) The car's position relative to the base of the cliff when the car lands in the ocean.
(b) The length of time the car is in the air.
2 answers
I'm not sure how to find a)