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 18.0 degree 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 2.90 m/s^2 for a distance of 35.0m to the edge of the cliff, which is 40.0 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

the information allows us to figure out 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 = accel = 2.90 m/s^2

d = distance = 35m

so the speed on leaving the cliff is:

vf^2 = 0 + 2(2.90)(35)

vf = 14.25

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

v(horizontal) = 14.25*cos18 = 13.55 m/s

v(vertical) = -14.25*sin18 = -4.40 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 - 1/2gt^2

y(t) = height at any time t,

y = initial position

v0y = initial y speed

y(t) = 40 - 4.40t - 4.9t^2

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

0 = 40 - 4.40t - 4.9t^2

this is a quadratic equation with solution

t = 2.44s

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 2.44 s is:

x = 13.55 m/s x 2.44 s = 33 m

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