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A major leaguer hits a baseball so that it leaves the bat at a speed of 31.5 and at an angle of 36.1above the horizontal. You can ignore air resistance.

At what two times is the baseball at a height of 8.00 above the point at which it left the bat?

  • physics -

    You need to give the dimensions of speed and height. I will assume they are m/s and meters

    Solve the equation

    Y = 8.00 = 31.5 t sin36.1 - 4.90 t^2
    8.00 = 18.56 t - 4.90 t^2

    The quadratic equation will provide two answers

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