Initial Velocity Question

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To determine the initial velocity of a water balloon launched vertically without a stopwatch, the discussion emphasizes using the conservation of energy and equations of motion. Key equations include VFy^2 = VOy^2 + 2a*Displacement Y, where the final velocity (Vfy) is zero at the peak height. Participants discuss measuring displacement with a ruler or protractor, suggesting using trigonometry to calculate height based on angles and distances. The conversation highlights the importance of understanding the relationship between height and velocity, noting that velocity increases with the square root of height changes. Overall, the thread provides insights into applying physics principles to solve the problem effectively.
MaNiFeST
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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
 
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Are you familiar with conservation of energy?
 
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, I am not familiar with conservation of energy
 
Ok, are you familiar with the equations of motion?
 
I am familiar with the equations for angles ( sin, cos), displacement in X,Y , etc,, for constant/non constant velocities and Trajectories
 
Look through them and try to find one that involves the parameters you think are going to be important and post it.
 
This isn't really a homework question but w/e

Thanks again
 
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
 
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Well, don't solve any equations yet, just look for some that might be useful, and well work from there.
 
  • #10
Would this work
VFy^2 = VOy^2 + 2a*Displacment Y
 
  • #11
Aha, you are on to something. Keep going.
 
  • #12
Displacment Y = {(Vo^2+sin(2*angle)} / g

Only other one i know that doesn't involve time
 
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  • #13
where did this come from? You were on the right track before. Maybe you should take a closer look at your first equation.
 
  • #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 won't have a stop watch when doing this experiment
 
  • #15
Yes, look at your first equation, and you tell me what each of those terms mean.
 
  • #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
 
  • #17
Ok, we need to get this terminology straight.

V_{fy} means the final velocity in the y direction.

V_{oy} means the initial velocity in the y direction.

Does this help at all?

Final Velocity of Y = is 0 m/s in this case

Yes, that's correct. Now when does this occur?
 
  • #18
VFy That occurs when the object stops right before it comes back down
I Know what the terminology means,
 
  • #19
MaNiFeST said:
VFy That occurs when the object stops right before it comes back down
I Know what the terminology means,

How would a ruler be useful given this information? What is the value of Vfy?
 
  • #20
Thats the thing, I have almost no idea
 
  • #21
Well, you just said it yourself:

VFy That occurs when the object stops right before it comes back down

What does that mean?
 
  • #22
It means that in the Y direction, the final velocity is 0 because there is no motion, it suspends in midair
 
  • #23
Exacttttttly. So what does that mean Vfy =?
 
  • #24
VFy = 0 m/s
 
  • #25
Bingo. Now, think about that ruler. What could you do with that ruler now that you have this new information?
 
  • #26
I can't measure the displacement of Y because it is too great, so I am really stumped
 
  • #27
Why not? Use a really big ruler. Make tick marks on the wall with chalk spaced half a foot apart it comes to that. If the velocity is small, do you think the distance is going to be great?
 
  • #28
Ooh i could gently launch it correct? nm , this wouldn't work , the force would be too low than when i launched it a second time
 
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  • #29
Sure, why not? You can launch it within reason. In fact, if you have a meter stick, calculate for me right now, what velocity it would need to reach 1 meter?

Let's see if it is resonable.
 
  • #30
I wish i could, i would, but all the supplies are at the lab in school :(
 

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