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.

(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.

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

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 !!

I get 35.9m for a, which checks out

but I keep getting 2.47s for b, which is supposedly wrong.

To find the car's position relative to the base of the cliff when it lands in the ocean, we need to find the horizontal and vertical distances the car travels.

First, let's find the horizontal distance the car travels. We can use the equation:

d = vt

Where d is the distance, v is the initial velocity, and t is the time. In this case, since the car starts from rest, the initial velocity is 0. The horizontal distance it travels is equal to the distance along the incline, which is 30.0 m.

Next, we need to find the vertical distance the car travels. We can use the equation:

d = vit + 1/2at²

Where d is the distance, vi is the initial vertical velocity, a is the acceleration, and t is the time. In this case, the car starts from rest vertically, so the initial vertical velocity is also 0. The acceleration, a, is equal to the constant acceleration down the incline, which is 3.21 m/s². We want to find the time it takes for the car to reach the edge of the cliff, so the final vertical distance is 50.0 m. Plugging in the values, the equation becomes:

50.0 = 0t + 1/2(3.21)t²

Simplifying:

50.0 = 1.605t²

Now, we can solve for t by rearranging the equation:

t² = 50.0 / 1.605

t ≈ √(50.0 / 1.605)

t ≈ 5.87 seconds

So, it takes approximately 5.87 seconds for the car to reach the edge of the cliff.

Now, to find the car's position relative to the base of the cliff when it lands in the ocean, we can use the equation:

d = vt

The horizontal distance traveled, d, is still 30.0 m. The time, t, is the same as the time it took for the car to reach the edge of the cliff, which is approximately 5.87 seconds. So, plugging in the values:

d = (0)t

d = 0

Therefore, the car's position relative to the base of the cliff when it lands in the ocean is 0 meters. In other words, the car lands right at the base of the cliff.

Now, to find the length of time the car is in the air, we can subtract the time it took for the car to reach the edge of the cliff from the total time it takes for the car to land in the ocean. The total time is approximately 5.87 seconds, and the time to reach the edge of the cliff is also 5.87 seconds. Therefore:

Length of time the car is in the air = Total time - Time to reach the edge of the cliff
= 5.87 seconds - 5.87 seconds
= 0 seconds

Therefore, the length of time the car is in the air is 0 seconds, as the car lands right at the base of the cliff.