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- Thread starter MaNiFeST
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- #2

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Are you familiar with conservation of energy?

- #3

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- #4

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Ok, are you familiar with the equations of motion?

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- #6

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This isn't really a homework question but w/e

Thanks again

Thanks again

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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

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

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- #10

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Would this work

VFy^2 = VOy^2 + 2a*Displacment Y

VFy^2 = VOy^2 + 2a*Displacment Y

- #11

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Aha, you are on to something. Keep going.

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Displacment Y = {(Vo^2+sin(2*angle)} / g

Only other one i know that doesn't involve time

Only other one i know that doesn't involve time

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- #13

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- #14

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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

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Yes, look at your first equation, and you tell me what each of those terms mean.

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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

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[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?

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I Know what the terminology means,

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MaNiFeST said:

I Know what the terminology means,

How would a ruler be useful given this information? What is the value of Vfy?

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Thats the thing, I have almost no idea

- #21

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

What does that mean?

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- #23

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Exacttttttly. So what does that mean Vfy =?

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VFy = 0 m/s

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I can't measure the displacement of Y because it is too great, so I am really stumped

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- #28

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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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Let's see if it is resonable.

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I wish i could, i would, but all the supplies are at the lab in school :(

- #31

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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

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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

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Yes, that is correct, in this equation you must make the acceleration -9.81m/s, do you know why?

- #34

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SO

4.429m/s = Vo * sin(90)

Vo = 4.429 m/s

- #35

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I must make a = -9.81 because it is going away and cannot have a negative velocity?

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