A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 20.0° 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 3.21 m/s2 for a distance of 30.0 m to the edge of the cliff, which is 50.0 m above the ocean.

Question

(a) Find the car's position relative to the base of the cliff when the car lands in the ocean.

(b) Find the length of time the car is in the air.

Answer 1

it seems to me that you have to do (b) first.

Use the acceleration rate and distance to calculate the car's velocity V as it leaves the cliff.
V = sqrt (2 a X)
It will leave with the same downward cliff angle as the cliff slope.
Get the horizontal and vertical components of the velocity as it leaves the cliff. Consider distance positive mesured downwards.
Vxo = V cos 20.0
Vyo = V sin 20.0
(b) Write an equation for the time T it takes to fall to the ocean.
Vyo*T + (1/2) gT^2 = 30.0 m.
(Solve for t)

(a) Horizontal position = Vxo*T

Answer 2

How fast is it going when it goes airborne?
v = a t
d = (1/2) a t^2
30 = .5 ( 3.21) t^2
t^2 = 18.6916
t = 4.32 s
v = a t = 3.21 * 4.32 = 13.9 m/s
call horizontal component u = 13.9 cos 20 = 13.04 m/s This is constant until we hit the bleak ocean
call vertical component Vo = 13.9 sin 20 = 4.75 m/s which is our starting speed down.
Do the vertical problem first to find how long it takes to fall 50 m with an initial speed down of 4.75 and an acceleration down of 9.8 m/s^2
x = Xo + Vo t + (1/2) a t^2
so
50 = 0 + 4.75 t + 4.9 t^2
4.9 t^2 + 4.75 t - 50 = 0
t = [ -4.75 ± sqrt (4.75^2 +4*4.9*50)]/9.8
=[-4.75 ± 31.7 ]/9.8 = 2.75 seconds in air
horizontal distance = u t = 13.04 * 2.75 = 35.9 meters
check my arithmetic !!

Answer 3

I get 35.9m for a, which checks out
but I keep getting 2.47s for b, which is supposedly wrong.