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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 = 13 m/s. The cliff is h = 58 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 - ,

    h=1/2 g t^2 solve for time t to hit the surface.
    speed? ?
    in the vertical v= at
    in the horizontal v=13m/s

    vf= sqrt(vvertical^2+13^2)

  • physics - ,

    Assume the initial kick is in a horizontal direction. Vertical motion is not affected.

    (1/2) g t^2 = 58 m when it hits the water

    Solve for t.

    (b) The new kinetic energy will be the initial kinetic energy PLUS M g H
    where H = 58 m.

    (M/2)*Vfinal^2 = (M/2)*vi^2 + M g H

    Cancel the M's and solve for Vfinal

    Vfinal^2 = vi^2 + 2 g H

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