Initial Velocity Question

  1. Nov 4, 2006 #1
    Ok So if I were to launch a waterballoon with a waterballoon launcher VERTICALLY , how could I find out the Initial Velocity With OUT a stop watch? My other supplies include a meter stick
     
  2. jcsd
  3. Nov 4, 2006 #2
    Are you familiar with conservation of energy?
     
  4. Nov 4, 2006 #3
    This is for a lab at school and you get Extra Credit if you can figure out the first part without a stop watch. Would i have to weigh it? And sorry, im not familiar with conservation of energy
     
  5. Nov 4, 2006 #4
    Ok, are you familiar with the equations of motion?
     
  6. Nov 4, 2006 #5
    I am familiar with the equations for angles ( sin, cos), displacement in X,Y , etc,, for constant/non constant velocities and Trajectories
     
  7. Nov 4, 2006 #6
    Look through them and try to find one that involves the parameters you think are going to be important and post it.
     
  8. Nov 4, 2006 #7
    This isnt really a homework question but w/e

    Thanks again
     
  9. Nov 4, 2006 #8
    Well if I am shooting an object vertically then Displacment X will be 0m, and
    Vy = Vo * sin(90) because the degrees will be 90
    VFy = 0 m/s

    Thats about all the variables i know atm

    Im unsure of which equation to use
     
    Last edited: Nov 4, 2006
  10. Nov 4, 2006 #9
    Well, dont solve any equations yet, just look for some that might be useful, and well work from there.
     
  11. Nov 4, 2006 #10
    Would this work
    VFy^2 = VOy^2 + 2a*Displacment Y
     
  12. Nov 4, 2006 #11
    Aha, you are on to something. Keep going.
     
  13. Nov 4, 2006 #12
    Displacment Y = {(Vo^2+sin(2*angle)} / g

    Only other one i know that doesnt involve time
     
    Last edited: Nov 4, 2006
  14. Nov 4, 2006 #13
    where did this come from? You were on the right track before. Maybe you should take a closer look at your first equation.
     
  15. Nov 4, 2006 #14
    its equation for X or Y displacment
    I am looking for Initial Velocity (Vo) and those two are the only ones that do not include time because i wont have a stop watch when doing this experiment
     
  16. Nov 4, 2006 #15
    Yes, look at your first equation, and you tell me what each of those terms mean.
     
  17. Nov 4, 2006 #16
    VFy^2 = VOy^2 + 2a*Displacment Y
    I am looking for Initial Velocity and not just VOy
    Final Velocity of Y = is 0 m/s in this case
    VOy^2 = is unknown
    a = 9.81m/s^2 in this case
    Displacment Y = is also unknown
     
  18. Nov 4, 2006 #17
    Ok, we need to get this terminology straight.

    [tex]V_{fy}[/tex] means the final velocity in the y direction.

    [tex] V_{oy} [/tex] means the initial velocity in the y direction.

    Does this help at all?

    Yes, that's correct. Now when does this occur?
     
  19. Nov 4, 2006 #18
    VFy That occurs when the object stops right before it comes back down
    I Know what the terminology means,
     
  20. Nov 4, 2006 #19
    How would a ruler be useful given this information? What is the value of Vfy?
     
  21. Nov 4, 2006 #20
    Thats the thing, I have almost no idea
     
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