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Vertical Kinematic Problem

  1. Mar 10, 2014 #1
    1. The problem statement, all variables and given/known data

    A stone is dropped of a cliff. A second stone is thrown down the cliff at 10 m/s at 0.5s after the first stone. When do the stones cross paths?

    2. Relevant equations

    Yf = vt + at^2/2

    3. The attempt at a solution

    My logic is the set the final position Yf equal for both stones and solve for time. Keeping in mind for the second stone that t = t + 0.5.

    After finding t, simply plug it into the equation to find Yf.

    Is this a valid method for solving this problem?
     
    Last edited: Mar 10, 2014
  2. jcsd
  3. Mar 10, 2014 #2

    SammyS

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    Almost.

    Two things:

    1. The second stone is in the air for 1/2 second less time than the first stone. You should decide what t stands for precisely.

    2. The question asks when the stone cross paths, not where .
     
  4. Mar 11, 2014 #3
    I would define t_total = t1 + t2 and then strictly use t2 in both of your equations:

    1) Figure out the value of Yf1 and v1 at t1 = 0.5 for the first object.
    2) Yf2 = 0 and v2 is given for the second object.
    3) Then set Yf1 = Yf2 both in terms of t2, and solve for t2 as you suggested.
    4) Give t_total = 0.5 + t2 as the final answer.
     
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