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

Question

A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 22.0° 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 3.52 m/s2 for a distance of 70.0 m to the edge of the cliff, which is 35.0 m above the ocean.

(a) Find the car's position relative to the base of the cliff when the car lands in the ocean.

(b) Find the length of time the car is in the air.

Answer keeps telling me I'm within 10% but cannot get it right perhaps due to roundoff error.

Explanation / Answer

here,
= 22
a = 3.52 m/s^2
D = 70 m
H = 35 m

using third equation of motion we have

Vf^2 – Vo^2 = 2 *a* d
Vf^2 – 0^2 = 2 * 3.52 * 70
Vf = 22.19 m/s

This is the velocity at the edge of the cliff :
Initial Vertical velocity = Voy = 22.19 * sin 22 = 8.219 m/s
Initial Horizontal velocity= Vox= 22.19 * cos 22 = 20.414 m/s

as the vertical velocity increases 9.8 m/s each second as it falls 35 m.
Initial Vertical velocity = 22.19 * sin 22 = 8.219 m/s
Acceleration = 9.8 m/s^2
Distance = 30 m

Final velocity^2 – 8.219^2 = 2 * 9.8 * 35 )
Final velocity = sqrt( (8.219)^2 + (2 * 9.8 * 35) )
Final velocity = 27.450 m/s

also
as,
Vf – Vo = a * t
27.450 – 8.219 = 9.8 * time
Time of fall = 1.962 seconds

During that 1.962 seconds, the car’s horizontal velocity = 20.414 m/s

Horizontal distance = 20.414 m/s * 1.962 s = 40.052 meters from the base of the cliff

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