Calculating Distance of Water Landing from Pool Hole | Torricelli's Theorem

[physhelp90]

Homework Statement

Instructor has not gone over this material and I want to know how to work this type of problem before she does. I don't even know how to begin.

A swimming pool is filled with water. It is 2.50 m tall and 3.00 m in diameter. There is a small 1.00 cm (in diameter) hole in the side of the pool and its 0.50 m below the top. How far from the pool will the water coming out of the hole land?

Homework Equations

The Attempt at a Solution

OK, here's my attempt using Torricelli's theorem, however I need to use Bernoulli's equation any help is appreciated.

Vx = Sqrt 2g(2.50 - 1.00)
Vx = 5.42 m/s This is the horizontal velocity.

t= sqrt 2h/g = .20 seconds = time water is in air

x = t x Vx

x = 1.11 meters ?

 

[Doc Al]

What have you tried so far?

Hint: Look up Torricelli's theorem.

 

[mgb_phys]

Bernoulli's equation will give you the sideways velocity of the water out of the hole.
Then it's just the normal vertical acceleration equation to find the time before it hits the ground (like a cannon ball fired horizontally)

 

[physhelp90]

What have you tried so far?

Hint: Look up Torricelli's theorem.

Here's my attempt using Torricelli's theorem but I need to use Bernoulli's equation

Vx = Sqrt 2g(2.50 - 1.00)
Vx = 5.42 m/s This is the horizontal velocity.

t= sqrt 2h/g = .20 seconds = time water is in air

x = t x Vx

x = 1.11 meters ?

Doesn't the diameter of the pool and the size of the hole have any play in the equation?

 

[Doc Al]

Here's my attempt using Torricelli's theorem but I need to use Bernoulli's equation

They are intimately related. (You can derive Torricelli's theorem from Bernoulli's theorem using certain simplifying assumptions.)

Vx = Sqrt 2g(2.50 - 1.00)
Vx = 5.42 m/s This is the horizontal velocity.

Where did you get those heights?

t= sqrt 2h/g = .20 seconds = time water is in air

Again, what height are you using?

Doesn't the diameter of the pool and the size of the hole have any play in the equation?

Not much. Compare the speed of water at the top compared to that at the hole.

 

[physhelp90]

Vx = Sqrt 2g(2.50 - 2.00)

Vx = 3.13 m/s horiz vel

3.13 x .41 = 1.28 m

Thanks for pointing out my calc in heights this is what I came up with based on 2.50 m = h0 2.00 m = h

How do I solve the same using Bernoulli's equation and is my answer of 1.28 m right?

 

[Doc Al]

Vx = Sqrt 2g(2.50 - 2.00)

Vx = 3.13 m/s horiz vel

Looks good.

3.13 x .41 = 1.28 m

Correct that value for time.

Thanks for pointing out my calc in heights this is what I came up with based on 2.50 m = h0 2.00 m = h

How do I solve the same using Bernoulli's equation?

Set up Bernoulli's equation to compare a point at the top of the pool to a point at the hole.

 

[physhelp90]

t= sqrt 2 x 2/9.8 = .64 s

so 3.13 x .64 = 2.0 m

 

[Doc Al]

t= sqrt 2 x 2/9.8 = .64 s

so 3.13 x .64 = 2.0 m

Looks good.

 

[physhelp90]

Bernoulli's equation will give you the sideways velocity of the water out of the hole.
Then it's just the normal vertical acceleration equation to find the time before it hits the ground (like a cannon ball fired horizontally)

Can you view my post again and determine if I came up with the same answer if I had used Bernoulli's equation and then calculated the time?

 

[Doc Al]

Can you view my post again and determine if I came up with the same answer if I had used Bernoulli's equation and then calculated the time?

If you apply Bernoulli's equation correctly, and make the same simplifying assumptions used in Torricelli's theorem, then you'll get the same answer. After all, that's where Torricelli's theorem comes from--it's just a special case of Bernoulli's equation.

Again, I urge you to actually look up Torricelli's theorem (in your book or on the web) and see how it's derived.