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Initial velocity, acceleration due to gravity question

  1. Sep 5, 2007 #1
    1. The problem statement, all variables and given/known data
    In this Lab you will toss water balloons over the edge of a raised sidewalk.
    The thrower should hold the balloon over the edge of the sidewalk and then toss it directly upward. As soon as the balloon is released the person on the ground should start the timer. Stop the timer when the balloon hits the ground. Balloon is released at 6.2m above ground and lands in 2 seconds. What is the initial velocity, final velocity and maximum height of travel?

    2. Relevant equations
    v=at+Vo
    y-y=Vot + 1/2at^2


    3. The attempt at a solution
    Tried to solve for Vo and y by substitution and I keep getting zero!
     
  2. jcsd
  3. Sep 5, 2007 #2
    "As soon as the balloon is released the person on the ground should start the timer."
    That should give you the initial velocity. From there, try solving again.
     
  4. Sep 5, 2007 #3
    use 0 for time and solve? that just gives me zero for initial velocity again.
     
  5. Sep 5, 2007 #4

    learningphysics

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    Your equation should be:

    y2 - y1 = v0t + (1/2)at^2

    solve for v0. you're given t = 2s
     
  6. Sep 5, 2007 #5
    Just to make sure I did this right, the initial velocity is 9.8m/s?
     
  7. Sep 5, 2007 #6

    learningphysics

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    No, that's not what I'm getting. Can you show your calculations? be careful about signs...
     
  8. Sep 5, 2007 #7
    0-6.2 = v0(2) + (1/2)-9.8(2)^2
    v0 = 6.7m/s
     
  9. Sep 5, 2007 #8

    learningphysics

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    That's right.
     
  10. Sep 5, 2007 #9
    great job helping me!! thanks
     
  11. Sep 5, 2007 #10

    learningphysics

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    no prob.
     
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