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Basic kinematics : 1-Dimensional collision in mid-air

  1. Aug 10, 2009 #1
    Hello. :smile:

    I've been stuck on this question for ages, and was hoping someone could help me with it.

    1. The problem statement, all variables and given/known data
    A stone is projected vertically upwards with a speed of 6m/s, and one second later a second stone is projected vertically upwards from the same point with the same speed.

    Find where the two stones meet.



    2. Relevant equations

    http://img193.imageshack.us/img193/4696/4bes.jpg [Broken]



    3. The attempt at a solution

    http://img195.imageshack.us/img195/908/4bp1.jpg [Broken]

    http://img140.imageshack.us/img140/1691/4bp2.jpg [Broken]

    http://img193.imageshack.us/img193/4383/4bp3.jpg [Broken]

    http://img148.imageshack.us/img148/6464/4bp4.jpg [Broken]

    As far as I can tell, I've worked out how high above the projection point the first stone is, when the second stone is launched.

    I also worked out that the first stone is travelling at 4m/s at 1m above the ground after 1 second.

    I tried doing this through trial and error, and came up with the collision at 0.15m above the projection point, 0.17 seconds after the second stone is launched, but I don't think that is the method I'm supposed to be using here. :shy:

    I don't think I've ever done a question like this before, so I would really appreciate any help. :smile:

    Thanks.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Aug 10, 2009 #2
    for first particle [tex]\[s_1 = ut_1 - \frac{1}{2}gt_1^2 \][/tex]
    for second particle [tex]\[s_2 = ut_2 - \frac{1}{2}gt_2^2 \][/tex]
    we know that [tex]\[t_1-t_2=1 , t_2=t_1-1\][/tex]
    then [tex]\[s_2 = u(t_1-1) - \frac{1}{2}g(t_1-1)^2 \][/tex].
    since they meet, s_1 should be equal to s_2.
    [tex]\[s_1 = ut_1 - \frac{1}{2}gt_1^2 = u(t_1-1) - \frac{1}{2}g(t_1-1)^2=s_2\][/tex]
    after solving the equation above for u=6m/s and g=10m/s^2, we find [tex]\[t_1=1.1s ,t_2=0.1s \][/tex] and [tex]\[s_1=s_2=0.55m\][/tex] .They meet at 0.55m above the ground.
     
  4. Aug 11, 2009 #3
    Thanks for the reply. :smile:

    Thank you so much for this. It's exactly what I was after. I haven't done anything like this before, so it took some time to get my head around it, but I think I understand it all now. I managed to work through it all and end up with 0.55 for s1 and s2.

    Thanks again. :smile:
     
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