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Vertically Thrown Stones

  1. Oct 5, 2013 #1
    1. The problem statement, all variables and given/known data
    A stone is thrown up vertically from the ground (the gravitational acceleration is g=10 m/s2). After a time Δt= 1 s, a second stone is thrown up vertically. The first stone has an initial speed v1= 7.0 m/s, and the second stone v2= 7.0 m/s.

    (a) At what time t after the first stone is thrown will the two stones be at the same altitude h above ground? (in seconds)

    (b) At what altitude h above ground will the two stones meet? (in meters)


    2. Relevant equations



    3. The attempt at a solution
    I created two equations : h=-.5gt^2+vt and h=-.5g(t-1)^2+v(t-1).
    Solved for t and got 1.2, then substituted t=1.2 in the equations and got h=1.2. Is that right?
     
  2. jcsd
  3. Oct 5, 2013 #2

    haruspex

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    Did you try substituting those times back into the equations to see if they both produce 1.2m? (I did, and they do :smile:)
     
  4. Oct 5, 2013 #3
    So am I right?
     
  5. Oct 5, 2013 #4

    haruspex

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    yes.
     
  6. Oct 6, 2013 #5
    A stone is thrown up vertically from the ground (the gravitational acceleration is g=10 m/s2). After a time Δt= 1 s, a second stone is thrown up vertically. The first stone has an initial speed v1= 13.0 m/s, and the second stone v2= 18.0 m/s.

    (a) At what time t after the first stone is thrown will the two stones be at the same altitude h above ground? (in seconds)
    (b) At what altitude h above ground will the two stones meet? (in meters)

    h=?

    Friends, i would really appreciate if you help me out to solve this question.
     
  7. Oct 6, 2013 #6
    hey, start a new/separate thread!!!
     
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