Would you like to formally "Introduce" yourself in the Introductions section, so we can all know more about you, where you live, your passions, etc.

]]>the position of x should be a function of t. x = p(t)

- can we use a cartesian co-ords? A water particle's position is

(x,y)= (x(t),y(t)) , in which x is the horizontal distance from bottle, y is the height it has fallen

define d= depth

x= √(2dg) t (1)

y= g t^2 /2 (2)

solving 2 for t gets

t= √(2y/g)

hence x can be expressed an explicit function of y as

x=2 √(dy)

if you place a full bottle on the ground, a hole at 1/n height of bottle and another hole at 1-1/n height will have their water stream intersect on the ground.

but the hole at the mid point will spray water further, with x=2d , the same as bottle height, when y= d = 1/2 bottle height

]]>Vectors don't have position, so you can place them anywhere you like in coordinate space, and as long as you don't altar their direction or magnitude, all is well.

That is completely true, in lower mathimatics. But when you get to upper level calculus, you start dealing with things such as vector fields, and there, vectors most certainly have a position.

]]>A vector is a quantity having both **magnitude** (meaning length) and **direction**. Think of a vector as an arrow drawn in coordinate space. Vectors don't have position, so you can place them anywhere you like in coordinate space, and as long as you don't altar their direction or magnitude, all is well.

Unit vectors are a convenient way to separate a vector into its component parts. A unit vector has magnitude 1 and is directed along one of the coordinate axes. The unit vector for x is called i, y is called j, and z is called k. They're written with a little hat (^) on top, like this: î (don't know if that'll come through).

So, a vector that has magnitude 5 and is directed 30 degrees above the horizontal has components (4)i and (3)j. If you add them together (using vector addition), they produce the final vector.

You can read more about them on your own. But, you don't need to know much about vectors to solve this problem. You can use the one equation to find the x position, and the other to find the y, and see if, at any one time, the three are the same. Like George, I'm too lazy to work it out for you, but it may involve systems of equations, graphing, and the like.

]]>* The water jets out with a certain horizontal velocity, which you could assume stays constant (ignoring air friction).

* The vertical velocity starts at 0, and increases by 9.81 m/s every second (ie g=9.81 m/s²) due to gravity.

BUT figuring out the horizontal velocity is the trick, and it depends mostly on the pressure of water (which increases with depth), but partly on the hole (friction mainly). George,Y's reference said the horizontal velocity would be Sqrt(2gh), but this may only be an approximation.

]]>Simply put: the position of x should be a function of t. x = p(t) what is p(t)? Or V(t) for velocity, and I can integrate to get position. Perhaps I need to get my hands on a physics book. :-/

]]>Thus a motion containing accelerations and velocities and displacements can be decomponented into 3 dimensions.

here 2-d is enough.

horizontally, no a, so v constantly √(2gh) thus displacement sx=v t, where sx stands for horizontal displacement.

vertically, a=g, v0=0 so vt = at = gt, displacement sy=1/2(v0+vt)t = 1/2 a t^2 = 1/2 g t^2

now we know after a certain time t, sy=d,

solve t out, t=√(2d/g)

sx=v t = √(2gh) √(2d/g)= 2√(hd)

sx=2√(hd), when sy=d

t is not contained in this formula, and it applies to water particle escaping from any holes. it states its horizontal displacement after having fallen distance d.

uh o, it seems a lot of difficulty to go further, lazy person like i would like to pass it onto next solver...

The position equation for a particle in free fall (where the position is r) is:

Note that r, v and g in this equation are vector quantities. For our case this is fairly simple to deal with because the initial velocity has only an x component and g has only a y component. So, the v term contributes to the x component of r, and the g term contributes to the y.

So, the x position at a given time t is given by:

And the y position at t is:

You should be able to use these to find the intersection. Have fun.

]]>Assuming you don't live on a really tall mountain or on the moon or anything, this can be taken as:

9.81 ms[sup]-2[/sup]

Edit because it looked weird before.

]]>Yeah Ryos this problem is as complex as we make it. But I'm not trying to land a rocket. What I'm most interested is, will the weight and the water and the height of the fall ballance out such that all the streams intersect in the same place.

sqrt (2gh) ? What does g stand for here?

]]>Sounds interesting.

What was the diameter of the holes, out of interest?]]>

This solution assumes that the potential energy due to pressure is the same as gravitational potential energy. It makes a kind of sense, because the fluid pressure in an open-air container arises due to gravity.

This problem can be about as complex as you want it to be. For example, as the water flows out of the tank, the fluid level falls, as does the pressure, so you have to take those rates into account. You can determine the flow rate based on the diameter of the hole and the pressure, but the flow rate *also* varies with depth.

The arc of the water is influenced by air resistance in a complex way.

Etc., etc., etc. You get the idea.

BTW, one useful little formula is P/ρ, or pressure/density, which gives the energy of a fluid under pressure per unit mass (J/kg).

Sorry if this is incoherent, my brain is fried from the physics test I just took.

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