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A lowly high diver pushes off horizontally with a speed of 2.75 m/s from the edge of a platform that is 10.0 m above the surface of the water. (a) At what horizontal distance from the edge of the platform is the diver 0.866 s after pushing off? (b) At what vertical distance above the surface of the water is the diver just then? (c) At what horizontal distance from the edge of the platform does the diver strike the water?

  • physics -

    a. d = Xo * t = 2.75m/s * 0.866s=2.38 m.

    b, d=0.5g*t^2 = 4.9*(0.866)^2 = 3.67 m.

    c. d = 0.5g*t^2 = 10 m.
    4.9t^2 = 10
    t^2 = 2.04
    Tf = 1.428 s. = Fall time.

    D = Xo * Tf = 2.75 * 1.428 = 3.93 m.

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