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Homework Help: Distance between two objects dropped at different times

  1. Sep 27, 2010 #1
    1. The problem statement, all variables and given/known data
    An object is dropped from the roof. A second object is dropped 1.50 s later. How far apart are the objects when the second one has reached a speed of 12.0 m/s.

    GIVEN
    Obj 1 - vi=0 m/s , a=-9.8 m/s/s
    Obj 2 - vi=0 m/s, vf=12.0 m/s, a=-9.8 m/s/s

    2. Relevant equations
    d=vf*t+1/2*a*t2


    3. The attempt at a solution
    d=12.0*1.50+1/2*-9.8*2.25
    d=18-11.025
    d=6.975

    I'm not sure exactly how to do this when there's two objects :S.
    The answer is 29.0 m, but I want to understand how to get it.
     
  2. jcsd
  3. Sep 27, 2010 #2

    ehild

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    Homework Helper

    Write the equations for the displacements of both objects. The usual equation is the distance travelled during the time since the object has been dropped: If you start a stopwatch when the first object is dropped, and the second one when the other one is dropped, the second watch will 1.5 s behind the first one.

    ehild
     
  4. Sep 27, 2010 #3
    So would it be:
    d=vf*t-1/2*a*tt
    d=12*1.5-1/2*-9.8*2.25
    d=18-(-11.025)
    d=29

    Oh I think I just messed up the sign on the equation.
    I think its right now.(?)
     
  5. Sep 27, 2010 #4

    jhae2.718

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    Gold Member

    You need to treat each object separately; i.e. you need an equation for the displacement of each object. (Technically these should be functions of time.)

    There is a time difference between the two objects; if you know the difference, how can you relate the times of the two objects? This relation will allow you to solve the first two equations.
     
  6. Sep 28, 2010 #5

    ehild

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    Homework Helper

    I do not know how you arrived to the equation, but it is correct.

    When the second object is dropped the first one has the velocity of 1.5 g and travelled 1/2 g 1.5^2 meters. After that, the distance travelled by it is
    s1=1/2 g 2.25+1.5gt + g/2 t^2, and that of the second object is s2=1/2gt. The distance between the objects is

    D=s1-s2 =1/2 g 2.25+1.5 gt.

    The second object reaches 12 m/s velocity in t=12/g time. Plugging back for t you get D=29.0 m.

    ehild
     
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