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Homework Help: Stone Throw, Found answer but don't understand how ?

  1. Oct 4, 2009 #1


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    1. The problem statement, all variables and given/known data
    A stone is thrown from a bridge with an initial vertical velocity of magnitude 4.0m/s/ Determine the stone's velocity after 2.2s if the direction of the initial velocity is (a) upward, and (b) downward. Neglect air resistance.

    2. Relevant equations
    Aav=Vf - Vi / Δt

    Vi + Aav(Δt) = Vf

    3. The attempt at a solution

    b) 2.6 x 101m/s
    Easy to find as you enter in the given information into the formula.

    (a) on the other hand I have difficulty understanding. I attempted the problem, with this formula
    (-Vi) + Aav(Δt) = Vf. The answer I got corresponds to the answer in text, but I don't really understand how that works. I'd like to know for similar problems in the future.
  2. jcsd
  3. Oct 4, 2009 #2
    You use the same equation to solve both problems, but if v_i is positive in b, it must be negative in problem a.
  4. Oct 4, 2009 #3


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    Yes that's apparently obvious, I want to understand why however. The "theory" behind it.
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