Water reach (Hose without and with a nozzle)

[Jenny Physics]

Homework Statement

A hose of inside diameter $1.5cm$ can reach a distance of $1.5m$. A nozzle is inserted in the hose and water can now reach $24m$. What is the inside diameter of the nozzle? The height is the same in both cases.

Homework Equations

Use continuity equation $v_{hose}A_{hose}=v_{nozzle}A_{nozzle}$

The Attempt at a Solution

Continuity equation gives $1.5/t\pi r_{hose}^{2}=24\pi r_{nozzle}^{2}$ which is the same as $1.5(0.015/2)^{2}=24r_{nozzle}^{2}$. So $r_{nozzle}=0.1875cm$ and the inside diameter will be twice that or $d=0.375cm$.

The solution however is $d=0.75cm$ (twice what I got). Where am I wrong?

 

[haruspex]

What is the relationship between launch speed and range?

 

[Jenny Physics]

What is the relationship between launch speed and range?

$x=v_{horizontal}t=v_{horizonta}\sqrt{\frac{2h}{g}}$. $h$ is the same in both cases

 

[haruspex]

h is the same in both cases

I see no reason why either t or h should be the same in both. What should we assume is the same?

 

[Jenny Physics]

I see no reason why either t or h should be the same in both. What should we assume is the same?

The problem states the height is the same in both cases (with and without the nozzle)

 

[haruspex]

The problem states the height is the same in both cases (with and without the nozzle)

I think they mean the heights of the end of the hose and of the point reached by the jet are the same.

 

[Jenny Physics]

I think they mean the heights of the end of the hose and of the point reached by the jet are the same.

Yes exactly

 

[haruspex]

Yes exactly

But you have used it in post #3 as though the height the water reaches at the top of its trajectory is the same in both scenarios. That is not the case.

 

[Jenny Physics]

But you have used it in post #3 as though the height the water reaches at the top of its trajectory is the same in both scenarios. That is not the case.

I see. Still if the water exits with only horizontal velocity it can only fall down not go up. That means the point the water reaches has to be at a lower height.

 

[haruspex]

if the water exits with only horizontal velocity

It says "can reach". What does that imply about the angle of the hose?

 

[Jenny Physics]

It says "can reach". What does that imply about the angle of the hose?

If the angle is 90 with the horizontal obviously can't reach. So some angle between 0 and 90 degrees. I am imagining it must be 45 for some reason but I don't see what reason

 

[gneill]

If the angle is 90 with the horizontal obviously can't reach. So some angle between 0 and 90 degrees. I am imagining it must be 45 for some reason but I don't see what reason

Look up the "range equation" for projectile motion.

 

[haruspex]

must be 45 for some reason but I don't see what reason

Because that gives the maximum range. But you do not need to assume it is 45 to solve the problem; you only have to assume that the angle does not change.