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March 30, 2017

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During a baseball game, a batter hits a popup to a fielder 64 m away.
The acceleration of gravity is 9.8 m/s
2
.
If the ball remains in the air for 6.7 s, how
high does it rise?
Answer in units of m

  • physics - ,

    Dx = Xo*6.7 = 64.
    Xo = 9.55 m/s = Hor. component of initial velocity.

    Tr = Tf = 6.7/2 = 3.35 s. = Rise and Fall time.

    Y = Yo + gt = 0.
    Yo - 9.8*3.35 = 0
    Yo = 32.83 m/s. = Ver. component of
    initial velocity.

    h = Yo*t + 0.5g*t^2.
    h = 32.83*3.35 -4.9*(3.35)^2 = 55 m.

  • physics - ,

    nun

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