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Physics

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A ball is thrown directly upward with an initial velocity of 12 m/s. If the ball is released from an initial height of 2.9 m above ground, how long is the ball in the air before landing on the ground? Ignore air drag.

  • Physics -

    height above ground =
    Y = 12 t - 4.9 t^2 + 2.9
    SWet that eq

  • Physics -

    height above ground =
    Y = 12 t -4.9 t^2 + 2.9 (meters)
    Set Y equal to zero and solve for t.
    4.9t^2 -12t -2.9 = 0
    There will be two solutions. Take the positive one.

    Use thye "quadratic formula":
    t = (1/9.8)[12 + sqrt(144 + 56.84)]
    = 2.671 s

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