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Kinematics, One Dimension, Acceleration, 2 objects problem

  1. Dec 4, 2017 #1
    1. The problem statement, all variables and given/known data
    Ball A is dropped from the top of a building of height h at the same instant that ball B is thrown vertically upward from the ground. At what height will the balls collide if the collision occurs when the balls are moving in the same direction and the speed of A is 4 times that of B.
    [​IMG]

    VA0 = 0 - the internal speed of ball A.
    h=h; a=g;
    tc - the time when the balls collide.
    VA(tc)=4*VB(tc)
    2. Relevant equations

    v = v(initial) +at
    x = x(initial) + v(initial)t + (at^2)/2

    3. The attempt at a solution
    XA(tc)=XB(tc)=hc (height whenn the balls collide)

    XA(tc)= h-(g*(tc)2)/2

    XB(tc)= XB(tb) - XB(tb→tc)=
    =VBO - (g*(tb)2)/2 - (g*(tc-tb)2)/2=
    =VBO - (g*(tb)2) - (g*(tc)2)/2 + g*tc*tb
    V(tb)=0 ⇒VB0=g*tb⇒
    ⇒XB(tc)=g*tc*tb - (g*(tc)2)/2

    XA(tc)=XB(tc)= h=g*tc*tb

    VA(tc)=4*VB(tc) ⇒ -g*tc = -g*(tc-tb) ⇒ 3tc= 4tb →witch is impossible!
     
  2. jcsd
  3. Dec 4, 2017 #2

    TSny

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    Gold Member

    Hello and welcome to PF!

    I think I follow your work up to the following
    I'm guessing that tb denotes the time at which B reaches maximum height. So, you are breaking up B's flight into two parts: the upward moving part and the downward moving part. It's not wrong to do this, but there is no need to break it up. The acceleration of B is constant throughout its entire flight. So, you can use the equation XB = VB0t - (g/2)t2 for the entire interval from t = 0 to t = tc.

    The mistake you made is in the second line quoted above where VBO should be multiplied by a time.
     
  4. Dec 4, 2017 #3
    thank you very much! I got the correct answer. <3
    p.s. sorry that i didn't clarified what tb means
     
  5. Dec 4, 2017 #4

    TSny

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    Gold Member

    OK. Good work.
     
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