Force of water jet hitting a wall at 13.6 m/s

[nibbel11]

but for now i only want to know how much force the wall gets when hit
so F=ma cause the mass is constant because there is no water loss or gain.
we are talking about a cylinder so
F=hr²πa
the hr²π is volume and the mass because it is water
is this the way to solve the force when the wall first has gotten hit

 

[theodoros.mihos]

$$ F = v\,\frac{dm}{dt} = \rho\,v\,\frac{dV}{dt} $$

 

[Simon Bridge]

so the only force there once was, was the moment the water hit the wall and then it stopped?

Only if the water stops at the wall. If the water bounces off with equal and opposite velocity, then there is twice as much change in momentum.

There is quite a lot missing from the question which will change how we interpret the information provided.
i.e. what is the context? How did this problem come to your attention?

i.e. if the context is a physics class on water flow in pipes - then I would have to consider that maybe the wall starts out flush against a pipe opening and the problem asks for the initial force on the wall (and you'd be best to use bernoulli then). But it could be that you are engaged in a home experiment involving pushing the vanes of a water-wheel with water from a hosepipe and you want an idea of what sort of acceleration you can expect, or it's a physics exercize in a lesson about Newton's laws.

 

[IgorIGP]

In principle, I agree with Simon Bridge. But theodoros.mihos is right that the velocity depends on the area ratio as long as the stream keeps continuity.The free jet behavior is quite a complex thing to make calculations without experimental data. As far as I know the experimentally based methods of calculating such things have those who deal with hydrotreating shipboard tanks (internal cleaning) and others.
I have questions on source data. Having the mass flow G=52,6L/s we mast have a nozzle with min diameter about 70mm. That means that in a pipe with D=240mm the velocity wil be more the 9 times lower. It can be said that the dynamical pressure (I do not sure what it means in English)) will be about zero. It is normaly fact. But the the expansion to 340mm may cause the practically full loss of velocity obtained in said nozzle.

 

[nibbel11]

simon i love games i made a website about what if things in games we in real life and with our rules. at this moment i look at a game called pokémon (creature fighting with attacks and stats etc) i am looking at a attack called hydropump. this is my most important source
i want to know what is the force delivered by this beam of water. btw look from 0,55 to 1,10
i know what everyone sais "it is a game things are inlogical" but that is why it is interesting for me. pleas help me even if it is completely against the law of physics i really need a way to calculate this

 

[Simon Bridge]

That is an impact/collision problem - use change in momentum.
A detailed workup of that situation would be quite complicated - you may be better working out the force based on the effect it has on different opponents.
Treat as a firehose.

 

[sophiecentaur]

simon i love games i made a website about what if things in games we in real life and with our rules. at this moment

If this is for a game simulation then momentum transfer calculations would be more than 'near enough' to make things look right, and relative motion can easily be included A game player can easily modify their reactions to cope with small departures from mechanical reality. Smooth dynamics and fine discrimination of input controls is far more important than the maths of the situation. No one is going to blame you when the simulation has tiny inaccuracies. How could they even spot them?