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A baseball player hits a home run over the left-field fence, which is 104 m from home plate. The ball is hit at a point 1.04 m directly above home plate, with an initial velocity directed 28.5° above the horizontal. By what distance does the baseball clear the 3.00 m high fence, if it passes over it 2.80 s after being hit?

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    The ball's horizontal velocity component is a constant 104/2.80 = 37.14 m/s. The initial vertical velocity component must be 37.14 tan28.5 = 20.17 m/s

    Next, compute the height Y of the ball 2.80 s after it is hit

    Y = 1.04 + 20.17*2.80 - (g/2)*(2.80)^2

    Subtract 3.0 m from that for the fence clearance distance.

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