Initial Velocity Question

In summary: Voy^2 + 2(-9.81m/s^2)(1m)So if the Voy is 4.429m for 1 m , then wouldn't it be different if the displacement was bigger? so how would i know how many times to multiply 4.429 by?If the displacement was bigger, then the velocity would be greater, and you would have to multiply by a larger number to reach the same final velocity.
  • #1
MaNiFeST
36
0
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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  • #2
Are you familiar with conservation of energy?
 
  • #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, I am not familiar with conservation of energy
 
  • #4
Ok, are you familiar with the equations of motion?
 
  • #5
I am familiar with the equations for angles ( sin, cos), displacement in X,Y , etc,, for constant/non constant velocities and Trajectories
 
  • #6
Look through them and try to find one that involves the parameters you think are going to be important and post it.
 
  • #7
This isn't really a homework question but w/e

Thanks again
 
  • #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
 
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  • #9
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.

[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?

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 :(
 
  • #31
What supplies, you have an equation!

Write the equation down for me! We just took care of all the terms!

Come on now, you know this. You were the one that supplied me with all the information.
 
  • #32
0 = Voy^2 + 2(-9.81m/s^2)(1m)
19.62=Voy^2
4.429m/s=Voy

! I just made a = -9.81 for going up so wouldn't end up with negative answer
 
  • #33
Yes, that is correct, in this equation you must make the acceleration -9.81m/s, do you know why?
 
  • #34
So if the Voy is 4.429m for 1 m , then wouldn't it be different if the displacement was bigger? so how would i know how many times to multiply 4.429 by?

SO

4.429m/s = Vo * sin(90)
Vo = 4.429 m/s
 
  • #35
I must make a = -9.81 because it is going away and cannot have a negative velocity?
 

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