How to find initial velocity with height, distance and angle

  • #1

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 dont know.
 

Answers and Replies

  • #2
haruspex
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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?
 

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