How to find initial velocity with height, distance and angle

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To find the initial velocity of a water fountain, the discussion emphasizes the need to analyze both vertical and horizontal components of motion using the provided height, distance, time, and angle. The equations mentioned include V = d/t for horizontal velocity and vf = vi + at for vertical motion, with considerations for acceleration due to gravity. Participants express frustration over multiple methods yielding different results, highlighting the complexity of the problem and the potential inaccuracies in the collected data. Clarification is sought regarding the definitions of time and height measurements, particularly whether the time refers to reaching maximum height or hitting the ground. Accurate interpretation of the data is crucial for determining the initial velocity effectively.
Jordann2222
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Homework Statement


We need to fin the initial velocity of a water fountain. The data I have collected is:
Height (bottom to apex)= 0.042 m
Distance (from spout to end on the stream)= 0.112 m
Time (it takes to travel)= .25 seconds
Angle the water shoots from= 50 degrees

Here is more if it helps.
Vertical
- D= .042 m
- T= .25 s
- Vfinal= 0 m/s
- a= 9.8 m/s2
- angle= 50 degrees
Horizontal
- D= .112 m
- T= .25 s
- angle= 50 degrees


Homework Equations


V= d/t, This can be used to find the horizontal component ( in the x direction) because Vxcomponent= average velocity.
Vycomponent= vertical initial velocity
vf= vi + at
d=Vi *t +1/2at2
a= acceleration t= time d=distance ( only used with Vertical data)
Vf2= vi2 + 2ad

The Attempt at a Solution


I know that I must find the x and y component vectors but I do not know what equation to use. I tried this four different ways and got four different answers and I'm just really frustrated. I was told to use the vf= vi + at equation for the vertical component and the v=d/t for the horizontal, and they said something about half time, I don't know.
 
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One problem is that you have collected more than enough information, creating more than one way to figure it out. Naturally, not all your data will be perfect, and the equations will ignore wind and drag. Consequently, different approaches will yield slightly different answers.
Is the angle to horizontal? I assume so.
Is the .25s the time to reach max height or the time to hit the ground? (What should be the relationship between the two?)
Is ground level the same as spout level? If not, is the .042m the height difference between them, the height from ground to max height of water, or the height from the spout to max height of water?
The datum I would trust least is the time measurement. As a check, if an object is dropped from a height of .042m, how long should it take to hit the ground?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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