Help with physics needed (hydrodynamics)

[DXM]

Hello all, I'm stuck on a very frustrating physics problem and I'm calling on the support of fellow Anandtechers.

Here's the problem:

A hole is punched at a height h in the side of a container of height H. The container is full of water. If the water is to shoot as far as possible horizontally, (a) how far from the bottom of the container should the hold be punched? (b) Neglecting frictional losses, how far (initially) from the side of the container will the water land?

Here's what I have so far:

Velocity of the water just as it exists the tank, v = sqrt(2g*(H - h)).
Using some kinematic equations I get t(ime) = sqrt(2gh).
The problem is, I can't for the life of me massage the figures to get the answer: (H / 2). Anyone have any ideas? Thanks in advance. 🙂

 

[DXM]

Anyone?

 

[DrPizza]

Originally posted by: DXM
Hello all, I'm stuck on a very frustrating physics problem and I'm calling on the support of fellow Anandtechers.

Here's the problem:

A hole is punched at a height h in the side of a container of height H. The container is full of water. If the water is to shoot as far as possible horizontally, (a) how far from the bottom of the container should the hold be punched? (b) Neglecting frictional losses, how far (initially) from the side of the container will the water land?

Here's what I have so far:

Velocity of the water just as it exists the tank, v = sqrt(2g*(H - h)).
Using some kinematic equations I get t(ime) = sqrt(2gh).
The problem is, I can't for the life of me massage the figures to get the answer: (H / 2). Anyone have any ideas? Thanks in advance. 🙂

Did you "massage" d = Vit + 1/2 at^2?
in which case, t = sqrt (2h/g)

 

[DXM]

Did you "massage" d = Vit + 1/2 at^2?
in which case, t = sqrt (2h/g)

Er yes, I meant t = sqrt( 2h/g ). Sorry for the typo. Any ideas?

 

[DrPizza]

Velocity of the water just as it exists the tank, v = sqrt(2g*(H - h)).
Using some kinematic equations I get t(ime) = sqrt(2gh).

So, you have v = sqrt(2g*(H-h)) and t = sqrt(2h/g)

distance = v*t = sqrt(2g*(H-h))*sqrt(2h/g) = sqrt(4h(H-h))
= sqrt(-4h^2 +4Hh)


You're trying to maximize the value of that function, so, you only need to maximize the inside part of the sqrt since the larger the number inside, the larger the value of the square root... maximize -4h^2 +4Hh
It's a parabola opening down, find the axis of symmetry... that's where it's maximized...
At H/2 (from the quadratic formula, use -b/2a)
Someplace I made a sign mistake [edit: duhhhhh, I fixed it] but it's too late for me to care... time to go home for dinner 🙂