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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 24.0° 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 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 car’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 = 20*cos(24) * 1.78
D = 32.5 m

Hope that makes sense!