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March 30, 2017

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A ball is projected horizontally from the edge of a table that is 1.15 m high, and it strikes the floor at a point 1.70 m from the base of the table.
(a) What is the initial speed of the ball?


m/s
(b) How high is the ball above the floor when its velocity vector makes a 45.0° angle with the horizontal?


m

  • physics - ,

    h = Vo*t + 0.5g*t^2 = 1.15m,
    0 + 4.9t^2 = 1.15,
    t^2 = 0.235,
    t = 0.484s = Time in flight.

    a. Dh = Xo * t = 1.70m, = Hor distance.
    0.484Xo = 1.70,
    Xo = 3.5m/s. = Initial hor velocity.

    b. tan45 = Y/Xo,
    Y = Xo*tan45 = 3.5 * 1 = 3.5m/s = ver.
    comp. of velocity.

    d = (Vf^2 - Vo^2) / 2g.
    d = ((3.5)^2 - 0) / 19.8 = 0.625m.
    h = 1.15 - 0.625 = 0.525m. above the floor.

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