Speed of water from a sprinkley

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Homework Help Overview

The problem involves calculating the velocity of water from lawn sprinkler heads and determining the flow rate based on the output diameter. The context includes projectile motion principles as the water exits the sprinkler heads at an angle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the horizontal distance covered by the water and the velocity, considering the projectile motion aspects. There are attempts to relate the area of water coverage to velocity and flow rate. Questions arise about how to derive time and velocity from the given parameters.

Discussion Status

Participants are actively exploring different methods to find the velocity and flow rate. Some hints and guidance have been offered regarding the use of projectile motion equations, but there is no explicit consensus on the final approach or solution yet.

Contextual Notes

There are noted constraints regarding missing information, specifically the initial and final vertical positions, which complicate the calculations. Participants express uncertainty about how to proceed with the variables involved.

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Homework Statement


Four lawn sprinkler heads are fed by a 1.9cm diameter pip.e The water comes out of the heads at angle of 35 degrees to the horizontal and covers a radius of 8m.
a) What is the velocity of the water coming out of each sprinkler head?
b) If the output diameter of each head is 3.0mm, how many liters of water do the four heads deliver per second?

Homework Equations


Continuity: A1*v1 = A2*v2

The Attempt at a Solution


For (a), I see the water covers a radius of 8m, so an area of 64pi meters. It must do this in a horizontal direction, so the speed is only in the X direction. I don't really know how to find out velocity though, because we don't know any other velocity.
For (b) I assume once you just multiply the area by the velocity found in (a)
 
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cashmoney805 said:
For (a), I see the water covers a radius of 8m, so an area of 64pi meters. It must do this in a horizontal direction, so the speed is only in the X direction. I don't really know how to find out velocity though, because we don't know any other velocity.
Hint: When the water leaves the sprinkler head, it becomes a projectile. (What if, instead of water, it was a ball throw at a given angle? How would you solve it then?)
For (b) I assume once you just multiply the area by the velocity found in (a)
Yes, the flow rate from each head is given by V*Area.
 
Doc Al said:
Hint: When the water leaves the sprinkler head, it becomes a projectile. (What if, instead of water, it was a ball throw at a given angle? How would you solve it then?)
I don't know what to do here. I know for a projectile you would figure out the total time in the air based on the y-component of velocity, then you can figure out the x component of velocity. So in this problem Vy = Vsin(35) and Vx=Vcos(35). How does this help though if I don't know V?
 
You're asked to find V. You know the distance traveled--use it.
 
Haha I don't know how to. There are 2 variables missing- V (which I have to find) and t. I would say get t from another equation, but I can't get it from the y=yo +Vyo + .5at^2 because y is missing.
 
cashmoney805 said:
There are 2 variables missing- V (which I have to find) and t.
Luckily you have two equations.
I would say get t from another equation, but I can't get it from the y=yo +Vyo + .5at^2 because y is missing.
What's the initial and final y position?
 
Is it .5at^2=Vsin35t, so t=Vsin(35)/4.9?

If so, I get V=83.4 which seems really really fast.
 
cashmoney805 said:
Is it .5at^2=Vsin35t, so t=Vsin(35)/4.9?
That's true.
If so, I get V=83.4 which seems really really fast.
That's not. Redo that last bit.
 
oh sqrt, so V= 9.13. Cool.

Just one more quick question, pretty related. I did a simple problem earlier today where it was like "Romeo wants to throw a ball so it hits Juliet's windows with only the x component of velocity." They gave you the angle which he threw it at.
In that problem they set Vyo=0 and solved for t based off the height provided. How could you set Vyo=though? If Vyo=0, wouldn't Vxo also =0?
As for this case, the reason Vyo isn't =0 is because the water is moving inside the pipe and comes up and out.
 

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