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Homework Help: Finding the vertical velocity

  1. Sep 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Alright we had a lab and we threw water balloons off the college's stadium to determine the height.In result the I got 9.2m from an average of 1.37sec using y=1/2gt^2

    Okay, this is where i get confused to use the right equation to determine vertical velocity.

    2. Relevant equations


    3. The attempt at a solution

    I did v=u+(9.8m/s^2)(2.81) to find the velocity. Is that right?
  2. jcsd
  3. Sep 30, 2012 #2


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    Staff: Mentor

    Hi Dejahbol, http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]
    You threw them? Or dropped them? Is 9.2m about right?
    I'm confused, too, because it's not clear whether you are trying to find u or v. Which?
    Last edited by a moderator: May 6, 2017
  4. Oct 1, 2012 #3
    Sorry i wasn't very elaborate on my information. The estimate height dropping the balloon at rest is 9.2 m. That was based on the average time of 1.37 seconds. For the second part of the lab we were given the choice to throw them down or up, and we decided with our group to throw it up and took 2.81 seconds to hit the ground. The question asks "Using the height of the building, find the vertical velocity with which you threw the balloon."

    And thanks for the welcome :)

    edit: I wasn't sure which kinematic equation to use for this particular part of lab.
  5. Oct 1, 2012 #4


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    Staff: Mentor

    You need an equation involving u as the only unknown, e.g., s = ut + ½at²

    (You'd be seeking initial velocity, u)
  6. Oct 1, 2012 #5
    So I end up doing this:


    = 2.3x10^-3.... i think i did something wrong lol.
  7. Oct 1, 2012 #6
    I've drawn it out for you below.

    Granted you threw the balloons straight forward as Vo and let them fall rather than launch them up and then down which would be an entirely different question. I've also solved it based on the premise that the information you gave in the OP is accurate so take a look.

  8. Oct 1, 2012 #7
    Wait i need vertical velocity :/. I got that part down and thanks for helping ;)
  9. Oct 1, 2012 #8


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    Staff: Mentor

    If it spent 2.81 secs in the air, why are you using 1.37 in the formula? :confused:
  10. Oct 1, 2012 #9
    didn't he say it spent 1.37 in the air?

    a bit confused now
  11. Oct 1, 2012 #10
    Thanks. I plugged in the wrong values... *sigh* 2.31. Sorry it's like 2 am and i've been up all day.
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