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Kinematics (rectilnear motion) simple

  1. Dec 22, 2013 #1
    Ball 1 is launched with an initial vertical velocity v1=160 ft/sec. Three seconds later, ball 2 is launched with an initial velocity v2. determine v2 if the balls are to collide at an altitude of 300ft.


    not sure where I am going wrong, do I have my limits correct?[URL=http://s1341.photobucket.com/user/nebula-314/media/20131222_191532_zpsf76c1ab5.jpg.html][PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/20131222_191532_zpsf76c1ab5.jpg[/URL][/PLAIN]
     
  2. jcsd
  3. Dec 22, 2013 #2
    v2 being initial velocity.
     
  4. Dec 22, 2013 #3
    I'm not sure what you're doing here, but if I were you I would use the following equation to calculate t for the first ball to reach 300ft, then using that t minus 3 (because the 2nd ball was launch 3 seconds later) and the same equation I would calculate v2.

    [itex]\Delta[/itex]x= [itex]\frac{1}{2}[/itex]a[itex]t^{2}[/itex]+[itex]v_{0}[/itex]t

    Just remember that here we take upside direction to be positive, hence a=-g and if you are measuring distance in ft you should use g=32.17 ft/s^2 , or convert ft to meter and use g=9.8 m/s^2

    Let me know how it goes :)
     
    Last edited: Dec 22, 2013
  5. Dec 23, 2013 #4
    Oh i was trying to derive that equation, and perhaps save some steps.
     
  6. Dec 23, 2013 #5
    But your solution did help a lot thanks.
     
  7. Dec 23, 2013 #6
    oh I see! I guess sometimes it's just easier to stick to the basics! :tongue2:

    No problem at all, happy to help
     
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