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Relative velocity

  1. Feb 23, 2013 #1
    1. The problem statement, all variables and given/known data

    Ball 1 is dropped from rest at a height of 10m above the ground. at the same time, ball 2 is thrown straight upward from ground level with an initial speed of 10m/s.

    2. Relevant equations


    find the relative velocity of the balls when they pass each other at t=1 second?


    equation given, relative velocity ==> v12= lv2-v1l

    vf=vi+at

    3. The attempt at a solution


    since ball1 is falling, in respect to earth, the acceleration is possitive thus, v1f=v1i+at equals, vf= 0+9.8m/s^2 *1s

    vf= 9.8 m/s

    for ball 2 the ball is thrown upwards, thus being in a negative acceleration in respect to earth thus,

    v2F=v2i+a*t
    V2f= 10m/s+ (-9.8m/s^2)(1s)

    v2f= 0.2

    now, lV2-V1l = 9.6m/s

    now, the correct answer is 10m/s

    and i wonder how come and why??

    could it be that in respect to each other, their acceleration is negative? thus, resulting in

    v1=-9.8
    v2=0.2

    lv2-v1l = l0.2+9.8l = 10m/s?
     
  2. jcsd
  3. Feb 23, 2013 #2

    G01

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    Remember that the balls are traveling in opposite directions so the velocities have opposite signs.
     
  4. Feb 23, 2013 #3
    so if i initially calculated that v1= 9.8 and v2= 0.2 their opposite signs would be -9.8 and -0.2 and when applied to the equation lv2-v1l = l -0.2+ 9.8l is still 9.6 m/s :-/

    and for some reason the professor uses, v(f) = vi-gt

    why is he subtracting gravitational force * time???????? shouldnt it be adding
     
    Last edited: Feb 23, 2013
  5. Feb 23, 2013 #4

    G01

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    You misinterpreted my above statement. If one ball is moving down and the other ball moves up then, v_up will have the opposite sign of v_down. Which one is positive and which is negative will depend on your choice of coordinate system.

    Your professor is probably using a convention where g is always positive. So, he changes the sign in the equations when necessary instead of the sign of g itself.

    Also, note that g is the acceleration due to gravity, not the force of gravity.
     
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