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Where will the stones meet?

  1. Apr 3, 2014 #1
    1. The problem statement, all variables and given/known data
    This is a very simple question. I even got the answer but I don't exactly I got it. Here it is:
    A stone is allowed to fall from the top of a 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where will the two stones meet.


    2. Relevant equations

    v2 - u2 = 2as

    s = ut + 1/2at2

    v = u + at
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 3, 2014 #2

    adjacent

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    It will meet when the distance from the ground of two stones become equal.
    Calculate that.
     
  4. Apr 3, 2014 #3
    We can also say that that they will meet at a point where the sum of distances covered by both the stones will 100. Is that correct?
     
  5. Apr 3, 2014 #4

    adjacent

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    No.Do it like this:

    How is the distance traveled by the thrown stone determined?
    That distance should be equal to the distance between the falling stone and the ground.So it's 100- the distance traveled by the falling stone.

    So you should use the equation relating displacement,time,initial velocity and acceleration(g in this case).
     
  6. Apr 3, 2014 #5
    Let A be the point from where the first stone is dropped. Let B be the point from where the second stone is projected. Let the balls meet at P after time t has elapsed. So, then AP + PB = 100. Let v and u be the speeds of the first and second stone respectively. So,
    1/2gt2 + ut - 1/2gt2 = 100
    So, ut = 100 and we get t = 4s.

    Is this correct?
     

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  7. Apr 3, 2014 #6
    It's OK. You can do it this way too.

    Actually this shows you that the two stones have a uniform relative motion.
    As they both move with the same acceleration, their relative acceleration is zero.
     
  8. Apr 3, 2014 #7
    @nasu Thanks. With this method I got the answer as 4s and distance 20m. Is the answer correct?
     
  9. Apr 4, 2014 #8

    adjacent

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    It's correct.
    My method:
    ##100-\frac{1}{2}gt^2=25t+\frac{1}{2} \times -gt^2## also gives t=4s s=20m.

    EDIT: I just realized that this is the same thing :shy:
     
  10. Apr 4, 2014 #9
    OK . Thanks.
     
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