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Constant Acceleration Question

  1. Jun 2, 2014 #1
    A rock is thrown vertically upward from ground level at time t = 0. At t = 1.6 s it passes the top of a tall tower, and 2.0 s later it reaches its maximum height. What is the height of the tower?

    My thought process:

    First find initial velocity:

    v=vo+at
    0=vo+(-9.8)(2)
    vo=19.6 m/s

    Then you can find the height of the tower, where x-xo is the displacement/height of the tower:
    x-xo=vo*t+.5*a*t^2
    x-xo=19.6*1.6+.5*-9.8*1.6^2
    x-xo=18.816 m

    However, WileyPlus says that I am wrong. What am I doing wrong?
     
  2. jcsd
  3. Jun 2, 2014 #2

    Nathanael

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    You did almost everything correctly. But is it true that it's maximum height is reached after 2 seconds?
    (That's what your math says.)
     
  4. Jun 2, 2014 #3
    OH, so I should have the maximum height as 3.6 seconds, correct?
     
  5. Jun 2, 2014 #4

    Nathanael

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    Correct.
     
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