posted by Anonymous .
A car is parked near a cliff overlooking the ocean on an incline that makes an angle of 29.1◦ with the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline and has a velocity 6 m/s when it reaches the edge of the cliff. The cliff is 42.3 m above the ocean.
The acceleration of gravity is 9.8 m/s2. How far is the car from the base of the cliff when the car hits the ocean? Answer in units of m.
The initial vertical velocity component is Vyi = -6.0 sin 29.1 = -2.918 m/s
The horizontal velocity component remains
Vx = 6.0 cos 29.1 = 5.243 m/s during the fall.
Solve this quadratic equation for the time T that it takes to fall to the ocean:
Y = 0 = 42.3 -2.918 T -(g/2) T^2
(Take the positive root; there will be two)
Once you have T, get the horizontal distance with
X = Vx * T