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A car rolling off a cliff

Q: A car is parked on a cliff overlooking the ocean on an incline that makes an angle of 24.0Â° 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 4.00 m/s2 for a distance of 50.0 m to the edge of the cliff. The cliff is 30.0 m above the ocean. Find (a) the carâ€™s position relative to the base of the cliff when the car lands in the ocean, and (b) the length of time the car is in the air.

A: You can use this equation to find Vf as the car leaves the cliff and begins to fall:

$2ad=V_f^2 - V_i^2$
$2(4)(50) = V_f^2 - 0$
$Vf = 20$

Now you can split this velocity up into its components. As the car leaves the cliff, it forms a triangle of velocities with 20 on the hypotenuse, and Vx = 20*cos(24), Vy = 20*sin(24)

These are the initial velocities in the horizontal and vertical directions, respectively.

Use $d = V_i*t + \frac{1}{2} at^2$ in the Vertical to solve for time.

$-30 = -20*sin(24) * t + 1/2 (-9.8) t^2$

Rearrange and use the quadratic formula, with a = -4/9, b = -20*sin(24) and c = 30

You should find that t = 1.78 seconds.

You can then use this time in D=VT in the horizontal, to solve for D.

D=VT
D = 20*cos(24) * 1.78
D = 32.5 m

Hope that makes sense!