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Problem solving?

  1. Jun 7, 2009 #1
    do tell me if the way i did this problem is right!!!

    q::An object is thrown vertically upward with a velocity of 30m/s.Four seconds later
    a second object is projected vertically with a velocity of 40 m/s .determine i)time(after first object is thrown)when two objects will meet each other in air ii) height from earth when two objects meet

    my probable solution:consider first object time as t and second object time as t+4,
    solve for t by equating their maximum height formula(s=ut+0.5gt^2) ?
  2. jcsd
  3. Jun 7, 2009 #2


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    But what makes you think the two objects will reach the same maximum height?
  4. Jun 7, 2009 #3
    why do you think max.height is not the same,as you throw the first object at time t
    and second ,time difference is t+4,or maybe iam wrong... please give me a solution then..
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