Bernoulli's equation

[notagenius08]

1. The problem statement, all variables and given/known data

A water tank of height h has a small hole at height y. The water is replenished to keep h from changing. The water squirting from the hole has range x. The range approaches zero as y goes to 0 because the water squirts right onto the table. The range also approaches zero as y goes to h because the horizontal velocity becomes zero. Thus there must be some height y between 0 and h for which the range is a maximum.

a)Find an algebraic expression for the flow speed v with which the water exits the hole at height y.

b)Find an algebraic expression for the range of a particle shot horizontally from height y with speed v.

c)Combine your expressions from parts A and B. Find the maximum range x_max

d)Find the height y of the hole.

2. Relevant equations

v1A1=v2A2
p1+.5(density)v1^2+(density)gy1=p2+.5(density)v2^2 +(density)gy2
deltaK+deltaU=Wext


3. The attempt at a solution

a)Solving for v2(assuming this is outside the can velocity):

v=sqrt of 2p1/denisty-2p2/density+v1^2+2gy1-2gy2

computer told me to check my answer, so I guess its wrong.

b)distance=rate*time

t=sqrt of 2h/g

x=v*t

haven't tried part c or d cause I cant get part a) right.

Thanks for any help.

[notagenius08]

Ok, so I have messed around a bunch and I think the answer for part a) is:

v=sqrt of 2*g(h-y)

Can anyone tell me if I am on the right track?

[bluebear19]

that's what i got for v, what did u get for the max range of x??

[bokonon]

I typed in your equation for v and its correct. For my services, you think you could explain a bit how you got it?

[bokonon]

Oh just realized that was posted some time, probably doesnt need my services, but . . .

Bluebear, I'm stuck at xmax = sqrt(4y(h-y)) I literally just substituted the answer from A into B.

But I think we have to get rid of y somehow so I'm hesitant to submit this. Have you gotten it yet?

[bluebear19]

the velociyty for an obj falling in kinematics is sqrt[2gh]
Vf^2 - Vi^2 = 2as where a = acceleration and s = displacement
here a = g since its falling and s = h-y = distance the water has fallen up to the point of the hole.
and Vi = 0
so Vf^2 = 2*g*(h-y)
and solve for v by taking the square root of both sides
and no I havent tried c yet but it is part of my homework so I will try very soon
hope this helps

[bluebear19]

the eqn you got i think only solves for x, but not Xmax

y = 1/2 h for max range, i think you can solve for the rest
you had the right idea with sqrt(4y(h-y)) take the 4 out and u get 2sqrt(y(h-y))
looking at inside the radical y(h-y). you want that value to be the largest as possible. But when you increase y, u decrease h-y and vice versa, so the biggest value u can get is 1/2 * 1/2. its a little common sense twist

[bokonon]

Great, thanks for your help. You havent by any chance also been assigned 15.58, the boat problem? There's a thread called "buoyancy force on a steel boat", where I describe my work on it.