Distance of throw of water jets

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To determine the distance of throw for a water jet in calm air, the key factors include the nozzle diameter, water speed at the nozzle outlet, and the angle of the nozzle. The trajectory of the water jet is influenced by air resistance, which can slow down the jet, and it is noted that the trajectory is not symmetrical, with the ascending part being longer than the descending side. For calculations, the exit speed and angle can be used in Newtonian motion equations to predict the throw distance. The equations provided are y = vTsin(θ) - g(T^2)/2 for vertical distance and x = vTcos(θ) for horizontal distance, where v is the exit velocity, T is time, θ is the exit angle, and g is the acceleration due to gravity. Understanding these dynamics is essential for accurate predictions, especially for applications like irrigation.
mauro
I need to determine the distance of throw of a water jet in calm air(no wind).
Is there anyone who has the equation ready as to say the throw T as a function of nozzle diameter, water speed at nozzle outlet (or pressure at nozzle inlet) and angle of the nozzle?


Thank you for any help you may have for me.

Bye
Mauro

PS How the air slow down a water jet? In case of a slug fired from a gun resistance is 1/2 C(ro)V^2 where ro is air density.
It is common experience to see that water jet trajectory is not simmetrical. The ascending part is much longer than descending side. Since I need to determine the distance of throw for a given jet, simmetrical solutions brings to grossly overestimation of the result.
Is there anyone with field experience, as for example nozzle manufacturers for irrigation purposes, willing to help me?
Or on the contrary a physics amateur with a theoretical formula?

Thank you to you all

Mauro
 
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If you have the exit speed and direction, its a simple Newtonian motion equation. You don't need anything else. You can also use pressure to calculate velocity if you need to.
 
like a projectile with the speed of the water at the exit...nothing else...
 
I think it is
y=vTsin(Θ)-g(T^2)/2
x=vTcos(Θ)
with v=exit velocity, T=time,Θ=exit angle above horizontal, and g=9.8 m/s^2.
 

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