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A person standing at the edge of a seaside cliff kicks a stone over the edge with a speed of vi = 20 m/s. The cliff is h = 53 m above the water's surface, as shown below.
(a) How long does it take for the stone to fall to the water?
(b) With what speed does it strike the water?

  • Physics -

    Since there is no initial vertical velocity component, the time to fall (T) is given by
    H = (1/2)gT^2
    H is the cliff height.
    T = sqrt(2H/g)

    The speed Vf at impact can be easily computed with conservation of energy.

    (1/2)m Vf^2 =(1/2)m Vo^2 = mgH
    The mass cancels out. Vo is the initial velocity (when kicked)

    Vf^2 = Vo^2 + 2 g H

  • Physics -

    Oh i get it. You have to use the equation for an accelerating straight-line object. Thanks

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