1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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:

    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:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook