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Ball drop question

  1. Oct 8, 2008 #1
    1. The problem statement, all variables and given/known data
    A ball of mass 0.35 kilograms is currently 1.1 meters off the floor and has a speed of 1.3 meters per second straight up. Assuming no energy is converted by air friction, how high will the ball go?


    2. Relevant equations
    PE=mgh
    KE=1/2mv^2


    3. The attempt at a solution
    Since there is no friction or work, i set the equation as mgh=mgx+1/2mv^2, solving for x(final height) but I get the wrong answer. And I'm pretty sure this is the right equation to use.
     
  2. jcsd
  3. Oct 8, 2008 #2
    yes I think that's a way to do it, maybe typo on calculator. I got 1.186m
     
  4. Oct 8, 2008 #3
    1.186? How did u get that? These are the values I plugged in.
    (.35*1.1*9.81)=(9.81*.35*x)+(.5*.35*1.3^2)
    From this x=(((.35*1.1*9.81)-(.(5*.35*1.3^2))/(9.81*.35)
    Using a Ti-83, and not rounding any answers, I get 1.014.

    Im confused...
     
  5. Oct 8, 2008 #4
    i think easier way is to use kinematics equation : vf^2=vi^2+2ad then add 1.1m with d to get it
     
  6. Oct 8, 2008 #5
    Yup that did it. Thanks! Though I'm still wondering why my equations didn't work. O well, thanks for help
     
  7. Oct 8, 2008 #6
    OMG nvm I'm a idiot. I can't do basic algebra -_____-, mine equation does work
     
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