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 25.0degree 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 3.97 m/s^2 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. m (b) Find the lenght of time the car is in t5he air. s

Explanation / Answer

we find that speed from:

vf^2=v0^2+2ad               where vf is final speed
v0=initial speed = 0
a = accel = 3.97 m/s/s
d=distance = 70m

so the speed on leaving the cliff is:

vf^2=0+2(3.97)(70)
vf^2 = 555.8
vf=23.57 m/s

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

v(horizontal) = 23.57 cos 25 = 21.366 m/s
v(vertical) = -23.57 sin 25 = -9.96 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)=35 -9.96t - 4.9t^2

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

0=35 -9.96t-4.9t^2

this is a quadratic equation with solution

t= 1.843 s............Ans. of part b

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 1.843 s is:

x=21.366 m/s x 1.843s =39.377 m........Ans. of part a

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