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Stone is thrown from the top of the building.

  1. Jan 29, 2014 #1
    1. The problem statement, all variables and given/known data
    A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above the ground. How much time elapses between the instant of release and the instant of impact with the ground?

    Vi = -20 m/s height = 60 m
    t = ?


    2. Relevant equations

    d = Vi.t + 1/2 a t2

    3. The attempt at a solution
    Couldn't figure out. How to use height as a distance?
     
  2. jcsd
  3. Jan 29, 2014 #2

    BvU

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    It has the same dimension ! The equations are OK.
     
  4. Jan 29, 2014 #3
    So 60 m = -20m/s . t + 1/2 a t2
    60 m = -20m/s t - 4.9 m/s2 t2?
    What should I do after this ...
     
  5. Jan 29, 2014 #4

    collinsmark

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    You seem to be using the convention that up is positive, which is fine. But is the stone traveling 60 m downward or upward? What does that tell you about the sign of the 60 m in your equation?

    Solve for t. (Hint: quadratic equation.)
     
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