A car is parked near a cliff overlooking the ocean on an incline that makes an a

Question

A car is parked near a cliff overlooking the
ocean on an incline that makes an angle of
20.3 degrees 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 5 m/s
when it reaches the edge of the cliff. The cliff
is 16.9 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 to 4 or 5 decimal places.

Explanation / Answer

s = (1/2)at^2 ... distance with constant acceleration as function of time 47.7 = (1/2)(5)t^2 t^2 = (2)(47.7)/5 = 15.05 t = 3.88 seconds to roll down the incline v = at = (6.34)(3.88) = 24.59 m/s u = horizontal velocity = (24.59)cos(28.5) = 21.61 m/s v = vertical velocity = (24.59)sin(28.5) = 11.73 m/s The time to fall can be found from: h = h0 - vt - (1/2)gt^2 ...... and h = 0 when the car lands 0 = 18.5 - (11.73)t - (1/2)(9.8)t^2 4.9t^2 + (11.73)t - 18.5 = 0 t = [-11.73 +/- SQRT(11.73^2 + 4*18.5*4.9)]/(2*4.9) t = [-11.73 + SQRT(362.6)]/9.8 t = [-11.43 + 19.04]/9.8 t = 0.777 seconds distance = (velocity)x(time) = u*t distance = (21.61)(0.777) = 16.79 meters The car will land 16.79 meters from the cliff

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