Posted by **Anonymous** on Sunday, November 8, 2009 at 7:05pm.

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**, Sunday, November 8, 2009 at 7:29pm
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**, Sunday, November 8, 2009 at 7:43pm
Thanks drwls

- physics -
**Anonymous**, Thursday, September 16, 2010 at 2:26pm
Awesome job.

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