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Physics

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A stone is thrown vertically upward with a speed of 12.0 m/s from the edge of a cliff 85.0 m high (Fig. 2-34).

(a) How much later does it reach the bottom of the cliff?
(b) What is its speed just before hitting?
(c) What total distance did it travel?

  • Physics -

    It is the same equations as always:

    d=Vi*t-1/2 g t^2
    Vf=Vi+gt

    Now for total distance, you will have to calculate the height it reaches at the top (when vf is zero). Double that, add to height of cliff.

  • Physics -

    5.05s

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