In: Physics
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 65.0 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 length of time the car is in the air. s
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:
vf2=v02+2ad
where vf is final speed
v0=initial speed = 0
a = accel = 2.75 m/s2
d=distance = 65m
so the speed on leaving the cliff is:
vf2=0+2(2.75)(65)
vf2 = 357.5
vf=18.91 m/s
now, we need to find the components of the car's velocity as it
leaves the cliff, they are:
v(horizontal) = 18.91 cos 17 =18.0837m/s
v(vertical) = -18.91 sin 17 = -5.528 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 gt2
y(t)=height at any time t,
y=initial position
v0y=initial y speed
so we have:
y(t)=35-5.528t-4.9t2
we want to find how long it takes for the car to reach y=0:
0=35-5.528t-4.9t2
this is a quadratic equation with solution
t= 2.167s
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 2.167 s is:
x=18.08m/s x 2.167s =39.187m