Fluid Dynamics Question(not too difficult)

[bob24]

Homework Statement

The spout heights for a container are 10 cm, 20cm, 30cm, and 40cm. The water level is maintained at a 45 cm height by an outside supply. Which water steam has the greatest range relative to the base of the container.
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I know that the one at 20cm is the greatest range already. I want to find out the actual range though.

Homework Equations

V=radical(2gh)

The Attempt at a Solution

For 20 cm: V=radical(2gh)
V=radical(2*9.81*.25m)
V=2.215m/s
So I have the velocity but I am confused about how to find the range. Which formula do I use and what values go where?

Do I use: Vfinal^2 - Vinitial^2=2aS, with S being distance
OR
Sfinal=1/2aT^2+ViT+Si with T=time and S=distance

 

[bob24]

UPDATE: Figured out the answer to my question, thanks :)

 

[nlightner]

I would suggest using the equation for range, which is R = V^2 sin(2θ)/g, where R is the range, V is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity. In this case, the initial velocity would be the velocity calculated for the 20 cm spout (2.215 m/s), the angle of projection would be 45 degrees (since the water is being projected horizontally), and the acceleration due to gravity would be 9.81 m/s^2. Plugging these values into the equation, we get:

R = (2.215 m/s)^2 * sin(2*45)/9.81 m/s^2
= 0.5 m

So the range for the 20 cm spout is 0.5 meters. This is the furthest distance that the water will travel from the base of the container.