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Motion problem with two rocks being released at different times

  1. Feb 25, 2008 #1
    1. The problem statement, all variables and given/known data
    the question is like this :you are in a balloon that is rising at a rate of 3.0m/s . you have two rocks. you release one when you are 50m above the ground. 2 seconds later you throw the other rock towards the ground . the two rocks hit the ground at the same time . what was the initial velocity of the second rock (with respect to the ground)


    2. Relevant equations
    we have to solve it only using the simple motion equations such as:
    ∆d = Vi t + ½ a t2



    3. The attempt at a solution
    what i did was to first say that :

    t(rock1)=t(rock2)-2s and we also know that the second rock is released when we are 56m above the ground

    then i proceeded to calculate the time it takes for the first rock to hit the ground (considering the fact that Vi for the first rock is 0)

    then i subtracted 2 from this time and used it to calculate Vi for the second rock
    i used ∆d = Vi t + ½ a t2 to calculate both these things

    the problem is im not sure im doing this right and i would appreciate it if someone could confirm my method
    thanks in advance
     
  2. jcsd
  3. Feb 25, 2008 #2

    Mentz114

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    Gold Member

    Surely -3 m/s ? The rocks will be moving with the balloon until they are dropped or thrown.

    Apart from that your method looks OK.
     
    Last edited: Feb 25, 2008
  4. Feb 25, 2008 #3
    oops i was thinking about a different problem
     
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