# Bernouilli's Theorem Question! Is my calculation correct?

1. Jul 23, 2010

### hcho88

1. The problem statement, all variables and given/known data
I just need someone to check my work.
(If what I'm doing is right, if not, tell me where I have gone wrong, and what I should do)
Please see attachment for the question.

2. Relevant equations
P + 0.5$$\rho$$v^2 + $$\rho$$gh = constant
$$\rho$$water = 1kg/m^3

3. The attempt at a solution
I'm not too sure if the pressure is different from inside the water tank to the outside.
The question is asking for the speed at which water leaves the spout, for is that the final speed?
I have assumed the pressure is the same inside and out, please tell me if I'm wrong.
The density varies with temperature, and the temperature inside the water tank and the temperature outside the water tank won't be the same, but I have assumed them to be the same, so the density will be the same on either side of the equation below.

P + 0.5$$\rho$$v^2 + $$\rho$$gh = P + 0.5$$\rho$$v'^2 + $$\rho$$gh'

Pressure cancels out, but the rest do not as v and h are not the same on each side.

0.5$$\rho$$v^2 + $$\rho$$gh = 0.5$$\rho$$v'^2 + $$\rho$$gh'
0.5v^2 + gh = 0.5v'^2 + gh'

I think the initial speed is 0, because I wouldn't know what other value it might be, again, please tell me if I'm wrong!

gh = 0.5v'^2 + gh'
9.8 x 12 = 0.5v'^2 + 9.8 x 4
78.4 = 0.5v'^2
156.8 = v'^2
So v' = 12.52 m/s

For the second part of the question, b)
isn't it just h=0?
For the water to hit the ground as far as possible from the base of the tank, the height at which the spout should be placed is 0m, isn't it?

#### Attached Files:

• ###### Q3.jpg
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2. Jul 23, 2010

### kuruman

Part (a) looks OK.

For part (b): Don't forget that the water drops vertically as soon as it is airborne. If it leaves the spout at ground level, it has zero distance to drop so it stays in the air for zero time, therefore it travels zero distance horizontally. Think "projectile motion."

3. Jul 23, 2010

### hcho88

So using projectile motion, like you said,
Maximum height = 12m
Hmax = (vsine$$\Theta$$)^2 / 2g
Maximum height is reached when $$\Theta$$=45 degrees.
So, v = 16.44 m/s

Time to reach ground:
t = 2v / g
t = 2(16.44) / 9.8
t = 3.36 s

So, displacement in Y direction:
Change in H = (initial velocity)t - 0.5gt^2
Change in H = 0.05 m = 5cm

Therefore can the water hit the ground as far as possible , 5 cm above the ground?
Is this correct???????????????

Last edited: Jul 23, 2010
4. Jul 24, 2010

### kuruman

No, it is not. You assume that you can vary the angle θ at which the water comes out. You cannot. The water comes out of the spout horizontally which means θ=0. All you can vary is the height h1 at which the spout is placed. First you need to find the range as a function of h1, then maximize the range with respect to h1.

5. Jul 24, 2010

### hcho88

Can I use the velocity I found (12.52 m/s) from part a) of the question?
Assuming I can...

theata = 0 degrees
horizontal speed = vcos(theata)
horizontal speed = 12.52 m/s
vertical speed = vsin(theata)
vertical speed = 0 m/s

H = (2(hsp x vsp)) / g
H = 0 ?
Hmax = vsp^2 / 2g
Hmax = 0 ?

I'm sorry :(
I don't understand..........
Could you please explain it a little more in detail?
Sorry..

6. Jul 24, 2010

### kuruman

You cannot use the velocity from part (a). That is the velocity at the specific height 4.0 m. You need to

1. Find an expression, not a number, for the velocity as a function of height h1.2. 2. Write an equation giving the horizontal position of the water at any time t.
3. Write an equation giving the vertical position at any time t.
4. Solve the vertical equation for the time of flight.
5. Use the time of flight to find the range.
6. Maximize the range.

Also, please try not to post everything in bold letters.