Can a Second Hole Be Added to a Leaking Tank to Achieve the Same Water Range?

[Okazaki]

Homework Statement

(This is a continuation of the problem where I proved Torricelli's Law: v = √(2gh))

The water level in a tank lies a distance H above the floor. There is a hole in the tank that a distance h below the water level
a.) Find the distance x from the wall of the tank at which the leaked stream of water hits the floor
b.) Could another hole be punched at another depth h' so that this second stream would have the same range? If so, at what depth?

Homework Equations

Δx = (v_f² - v_i²)/g
v = √(2gh)

The Attempt at a Solution

At first, I was not really sure what to do here at all (and I probably ended up solving it completely wrong.)

So, basically, I wasn't exactly sure how to find the distance, since I only knew the initial velocity in the x direction (or, at least, I assumed v = √(2gh)) was the velocity of the water in the x direction based off the picture in the problem.)

So, what I tried was:
(Since the equation I had to derive from Torricelli's law in the problem I had to do before this ended up being Δv = √(2gh), I assumed:)

Δx = (2gh)/g
= 2h

For b, I'm completely confused on even where to begin solving for the equation.

 

[mfb]

Where does that Δx formula come from?
Your water, once it leaves the tank, moves horizontally and is in free fall. What do you know about objects in free fall?

 

[Okazaki]

Where does that Δx formula come from?
Your water, once it leaves the tank, moves horizontally and is in free fall. What do you know about objects in free fall?

Wait a minute...the water has a v_y when it leaves the tank of 0 m/s, doesn't it? If so, then I think I can solve it.

 

[mfb]

Right, assuming "y" is your vertical coordinate.

 

[Okazaki]

Right, assuming "y" is your vertical coordinate.

Well, I over-thought that problem.

Then:

v_i-y = 0 m/s
So:
==> v_f-y = v_i-y + 2gh
= 0 + 2gh
= 2gh

So:
v_f-y = v_i-y + gt
t = v_f-y/g
==> t = √(2gh)/g

So:
This is where I use: v = √(2gh)
Δx = v_i-yt + 0.5at²
= √(2g*(H-h)) * √(2gh)/g
= 2 √(h(H-h))

Still not sure about b, though. Would I just work backwards?

 

[haruspex]

Δx = v_i-yt + 0.5at²
= √(2g*(H-h)) * √(2gh)/g
= 2 √(h(H-h))

Still not sure about b, though. Would I just work backwards?

If a hole at height h' produces the same Δx, what equation can you write?

 

[mfb]

A velocity cannot be equal to 2gh, the units do not match. And I don't understand the other steps, but the result looks right.

 

[Okazaki]

A velocity cannot be equal to 2gh, the units do not match. And I don't understand the other steps, but the result looks right.

v = √(2gh) was what I had to prove in the previous problem.

 

[Okazaki]

A velocity cannot be equal to 2gh, the units do not match. And I don't understand the other steps, but the result looks right.

Wait, I see the issue. In my notes, I think I had a square root sign. It just never made it onto paper.

 

[Okazaki]

If a hole at height h' produces the same Δx, what equation can you write?

Well, if it's at h', then the V_i-x will be √(2g(H-h')) and you can work backwards from there.

 

[haruspex]

Well, if it's at h', then the V_i-x will be √(2g(H-h')) and you can work backwards from there.

Better, consider h' in respect of this equation in your post #5: Δx = 2 √(h(H-h)).

 

[mfb]

Drawing a sketch of Δx as function of h could be useful.