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A car is parked near a cliff overlooking the ocean on an incline that makes an angle of 29.1◦ with the horizontal. The negligent driver leaves the car in neutral, and the emergency brakes are defective. The car rolls from rest down the incline and has a velocity 6 m/s when it reaches the edge of the cliff. The cliff is 42.3 m above the ocean.
The acceleration of gravity is 9.8 m/s2. How far is the car from the base of the cliff when the car hits the ocean? Answer in units of m.

  • physics -

    The initial vertical velocity component is Vyi = -6.0 sin 29.1 = -2.918 m/s

    The horizontal velocity component remains
    Vx = 6.0 cos 29.1 = 5.243 m/s during the fall.

    Solve this quadratic equation for the time T that it takes to fall to the ocean:

    Y = 0 = 42.3 -2.918 T -(g/2) T^2
    (Take the positive root; there will be two)

    Once you have T, get the horizontal distance with
    X = Vx * T

  • physics -

    Thanks drwls

  • physics -

    Awesome job.

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