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 19.0° 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 3.06 m/s2 for a distance of 40.0 m to the edge of the cliff, which is 45.0 m above the ocean. (a) Find the car's position relative to the base of the cliff when the car lands in the ocean. m (b) Find the length of time the car is in the air. s

Explanation / Answer

first we find the horizontal and vertical speed of the car when it goes off the edge of the cliff

we find that speed from

vf^2 = v0^2 + 2ad where vf is final speed

v0=initial speed = 0

a = acceleration = 3.06 m/s^2

d = distance = 40m

so the speed on leaving the cliff is

vf^2 = 0 + 2*(3.06)*(40)

vf^2 = 244.8

vf = 15.64 m/s

now, we need to find the components of the car's velocity as it leaves the cliff, they are

v(horizontal) = 15.64*cos19 = 14.78 m/s

v(vertical) = -16.35 sin 24 = - 5.09 m/s

we need to find the time the car is in the air, for this we use the equation of motion

y(t)=y0 + v0y*t - 1/2*gt^2

y(t)=height at any time t, y0=initial position

v0y=initial y speed

so we have

y(t) = 45 - 5.09*t - 4.9*t^2

we want to find how long it takes for the car to reach y=0:

0 = 45 - 5.09*t - 4.9*t^2

this is a quadratic equation with solution

t= 2.55 sec

since the horizontal speed will not change once the car leaves the clilff (since there are no horizontal forces acting), we have that the horizontal distance traveled in 2.55 sec is

x = 14.78 m/s * 2.55 s = 37.69 m

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