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Initial Velocity Question

by MaNiFeST
Tags: initial, velocity
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MaNiFeST
#1
Nov4-06, 06:10 PM
P: 36
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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Cyrus
#2
Nov4-06, 06:11 PM
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Are you familiar with conservation of energy?
MaNiFeST
#3
Nov4-06, 06:13 PM
P: 36
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

Cyrus
#4
Nov4-06, 06:15 PM
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Initial Velocity Question

Ok, are you familiar with the equations of motion?
MaNiFeST
#5
Nov4-06, 06:17 PM
P: 36
I am familiar with the equations for angles ( sin, cos), displacement in X,Y , etc,, for constant/non constant velocities and Trajectories
Cyrus
#6
Nov4-06, 06:19 PM
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P: 4,777
Look through them and try to find one that involves the parameters you think are going to be important and post it.
MaNiFeST
#7
Nov4-06, 06:19 PM
P: 36
This isnt really a homework question but w/e

Thanks again
MaNiFeST
#8
Nov4-06, 06:22 PM
P: 36
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
Cyrus
#9
Nov4-06, 06:25 PM
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P: 4,777
Well, dont solve any equations yet, just look for some that might be useful, and well work from there.
MaNiFeST
#10
Nov4-06, 06:27 PM
P: 36
Would this work
VFy^2 = VOy^2 + 2a*Displacment Y
Cyrus
#11
Nov4-06, 06:28 PM
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P: 4,777
Aha, you are on to something. Keep going.
MaNiFeST
#12
Nov4-06, 06:30 PM
P: 36
Displacment Y = {(Vo^2+sin(2*angle)} / g

Only other one i know that doesnt involve time
Cyrus
#13
Nov4-06, 06:33 PM
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P: 4,777
where did this come from? You were on the right track before. Maybe you should take a closer look at your first equation.
MaNiFeST
#14
Nov4-06, 06:35 PM
P: 36
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
Cyrus
#15
Nov4-06, 06:37 PM
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P: 4,777
Yes, look at your first equation, and you tell me what each of those terms mean.
MaNiFeST
#16
Nov4-06, 06:39 PM
P: 36
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
Cyrus
#17
Nov4-06, 06:41 PM
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P: 4,777
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?
MaNiFeST
#18
Nov4-06, 06:44 PM
P: 36
VFy That occurs when the object stops right before it comes back down
I Know what the terminology means,


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