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Posted by on Saturday, October 1, 2011 at 11:03pm.

During a baseball game, a batter hits a pop-up to a fielder 83 m away.
The acceleration of gravity is 9.8 m/s^2.
If the ball remains in the air for 6.5 s, how high does it rise?
Answer in units of m

  • Physics - , Sunday, October 2, 2011 at 11:28am

    )A batter hits a pop-up to a fielder 83 m away. If the ball remains in the air for 6.5 s, how high does it rise?

    Rise time = fall time = 3.25sec.
    Vf = 0 = Vo - 9.8(3.25) or Vo = 31.85m/s

    d = Vo^2(sin(2µ))/g
    ----31.85^2(sin(2µ))/9.8 or µ = 26.65º

    Then, height h = Vo^2(sin^(x)/2g or
    h = 31.85^2(sin(26.65))/2(9.8) = 10.41m

  • Physics - , Monday, November 2, 2015 at 8:48pm

    dafuq

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