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Standing on the top ledge of a 55 meter high ?

  1. Jan 19, 2010 #1
    Standing on the top ledge of a 55 meter high building you throw a ball straight up with an initial speed of 29 m/s. How long, to the nearest second, does it take to hit the ground?

    If the building in the previous question was 56 meters high and you throw the ball at 30 m/s, how high to the nearest meter does it go?

    The correct answer to problem number one is 7
    And number two is 102

    Can't figure the formulas that need to be used in order to solve these..
    Any help would be appreciated...
  2. jcsd
  3. Jan 19, 2010 #2


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    Science Advisor
    Gold Member

    General formula:

    d=h + vt + gt2/2.
    where d is height at time t.

    h=initial height
    v=initial speed (+ is up)
    g=gravity constant (use - since it is downward) = (approx) 9.8 meters/sec2

    When object hits the ground d=0, you need to solve for t to get time.

    s(the speed at time t) = v+gt. To find max height, find t when s=0 and compute d for this t.
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