# physics

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 36� 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 5.17 m/s2 and travels 27.3 m to the edge of the cliff. The cliff is 33.7 m above the ocean. The acceleration of gravity is 9.8 m/s2.

Find the car’s position relative to the base of the cliff when the car lands in the ocean. Answer in units of m

Find the length of time the car is in the air. Answer in units of s

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1. Just plug in your basic equations of motion:

v = at
s = 1/2 at^2

During the time the car was rolling, it traveled

s = 27.3m to the edge of the cliff. So,

27.3 = 1/2 * 5.17 t^2
t^2 = 10.56
t = 3.25 sec

So, when the car rolls off the cliff,
v = at = 5.17 * 3.25 = 16.8m/s
Since the ground is sloping, we need the horizontal/vertical components. The 36 degrees is handy, as we have a 3-4-5 triangle.
vh = 3/5 * 16.8 = 10.08 m/s
vv = 4/5 * 16.8 = 13.44 m/s

Now, what happens while the car is falling? Its vertical acceleration is just gravity.

33.7 = 4.9 t^2
t^2 = 6.877
t = 2.622 sec

So, what happens during those 6.22 seconds? The horizontal velocity does not change, so

vh = 10.08m/s for 2.622 sec,
s = vt = 10.08 * 2.622 = 26.43m from the cliff.

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