Posted by **John** on Wednesday, October 23, 2013 at 10:10pm.

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 22◦ 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 4.3 m/s2 and travels 43 m to the edge of the cliff. The cliff is 33 m above the ocean.

How long is the car in the air? The acceler- ation of gravity is 9.81 m/s2 .

Answer in units of s

What is the carâ€™s position relative to the base of the cliff when the car lands in the ocean?

Answer in units of m

- Physics -
**Henry**, Friday, October 25, 2013 at 8:46pm
V^2 = Vo^2 + 2a*d

V^2 = 0 + 8.6*43 = 369.8

V = 19.23 m/s At the edge of ihe cliff.

Vo = 19.23m/s[-22o] At the edge of the cliff.

Xo = 19.23*cos(-22) = 17.83 m/s.

Yo = 19.23*sin(-22)=-7.20 m/s.=7.20 m/s,

downward.

a. h = Yo*t + 0.5g*t^2 = 33 m.

4.9t^2 + 7.2t - 33 = 0

Use Quadratic Formula.

t = 1.96 s. In air.

b. Dx=Xo * t = 17.83m/s * 1.96s=34.9 m

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