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General Problem

  1. Sep 7, 2008 #1
    I've been kicking myself in the head all day trying to figure this one out. I'm completely bewildered.:uhh: Please help meeee!


    A stone is thrown vertically upward at a speed of 43.00 m/s at time t=0. A second stone is thrown upward with the same speed 2.250 seconds later. At what time are the two stones at the same height?

    AND:

    At what height do the two stones pass each other?

    *Just need some help setting it up. I've never done an equation comparing two objects before.*
     
    Last edited: Sep 7, 2008
  2. jcsd
  3. Sep 7, 2008 #2

    atyy

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    For one object you can write an equation relating its height h1, and the time elapsed from the time it was thrown t1. You can write the same sort of equation relating h2 and t2. From here you have two equations and four unknowns (h1,t1,h2,t2)

    A second stone is thrown upward with the same speed 2.250 seconds later means that t2=t1+2.25s. This is your third equation.

    Think about the mathematical relationship between h1 and h2 when the two stones are at the same height. This should give you your fourth equation.

    You will now have 4 equations and 4 unknowns, which means you can have enough equations to deduce all the unknowns.
     
  4. Sep 7, 2008 #3


    I'm not sure I'm totally following you. Both H1 and H2 will be the same since that corresponds to the maximum height they reach, which I got to be 94.34 meters. I also calculated the time it took the first stone thrown to go up and back down to where it started from. t1=0 (starting point) & t1=8.78 seconds (finish point).


    EDIT: I just solved at what time the two stones are at the same height, which is 5.5 seconds, by drawing a picture and working from there. I'm still having trouble trying to get the height where they're both at the same height.
     
    Last edited: Sep 7, 2008
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