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 17.0° 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 2.75 m/s2 for a distance of 45.0 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. Incorrect: Your answer is incorrect. Your response differs from the correct answer by more than 10%. Double check your calculations. m (b) Find the length of time the car is in the air.

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 = 2.75 m/s/s
d=distance = 45m

so the speed on leaving the cliff is:

vf^2=0+2(2.75)(45)
vf^2 = 247.5
vf=15.73 m/s

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

v(horizontal) = 15.73 cos 17 = 15.04m/s
v(vertical) = -15.73 sin 17 = -4.59 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, y==initial position
v0y=initial y speed

so we have:

y(t)=45 -4.59t -4.9t^2

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

0=45-4.59t-4.9t^2

this is a quadratic equation with solution

t= 2.59s

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 in2.59 s is:

x=15.04m/s x2.59s =38.95m

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