Posted by Anonymous on .
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.

physics 
drwls,
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 
physics 
Anonymous,
Thanks drwls

physics 
Anonymous,
Awesome job.