[sheila1983]

I'm trying to solve the following problem for the water tower.
A large water tower has been built on top of a steep hill. The water is filled from y = 0 to y = h metres and the distance h = 5 metres (this would be the depth of water in the tower).
If a hole is punched into the side of the water tower at y = 1 metre, what is the pressure just inside the hole and what is the velocity of the water as it exits the hole.
Also, if the angle of the hill is 45 degrees, how long from the time the water exits the hole does it take to hit the hill and what are the (x,y) coordinates of where the water hits the hill.
Finally, what is the velocity of water when it hits the hill (in vector form).
Thank you in advance to anyone who can help.

[ebaines]

Hint: The pressue of water at depth 'h' is . The velocity of water going through the hole at depth h can be found using energy principles, and turns out to be the same as the velocity of a stone being dropped from height h. The rest of the problem is a simple projectile trajectory calculation.

[sheila1983]

Hint: The pressue of water at depth 'h' is . The velocity of water going through the hole at depth h can be found using energy principles, and turns out to be the same as the velocity of a stone being dropped from height h. The rest of the problem is a simple projectile trajectory calculation.

OK, thanks, so I calculated the water pressure using pressure = liquid density x gravity x water depth = (1000 kg m-3 x 9.8 ms-2 x 4m) = 39, 200 Pascals. (I got 4 metres by subtracting 1 metre from 5 metres).
Then, I applied the Law of Conservation of Energy, using (1/2 mv^2 + mgh)before = (1/2 mv^2 + mgh)after.
I then assumed that the water isn't moving before, so (1/2 mv^2) before = 0 and, also, as the water exits the hole, all of its potential energy is gone, so (mgh)after = 0. Thus, the velocity of the water as it exits the hole would be the square root of 2gh = 2(9.8 ms-2)(4metres) = 8.85 ms-1.
For part (c), I am puzzled. Here is why. If the pressure of the water is low when exiting the hole, it might just flow down the side of the tower and you could possibly work out the time the water exits the hole until it hits the top of the hill, using v = u + at, where v = final velocity of the water when it hits the hill (0 ms-1) and u is its initial velocity.
However, why would you be give the angle of the hill then?
If the pressure of the water is high, it might shoot out the tower at a right angle, so in that case, if you were using the equations for projectile motion, wouldn't the angle of launch (theta) be equal to zero degrees? Also, does the angle of exit of the water from the hole have any relationship with the angle of the hill?

[ebaines]

Assume the water shoots out horizontally. Imagine a water droplet acting like a projectile from the hole with initial horizontal velocity as you have already calculated and no initial vertical component . Treat just like any other projctile problem you 've done in the past.

[sheila1983]

Assume the water shoots out horizontally. Imagine a water droplet acting like a projectile from the hole with initial horizontal velocity as you have already calculated and no initial vertical component . Treat just like any other projctile problem you 've done in the past.

OK, thanks, the diagram helps a lot.
So, to work out how long from the time the water exits the hole it takes to hit the hill, I firstly calculated the maximum distance where the water hits the hill on the x axis, using one of the standard equations of motion for a projectile.
Since the water is initially only moving in the x direction, then v sine theta (where v is the initial velocity of water as it exits the hill) will be zero and the standard equation will reduce to: x(max.) = v / g times the square root of 2 times g times h = (8.85 ms-1 / 9.8 ms-2) times (2 x 9.8 ms-2 x 1 metre) = 4 metres.
Then, since the angle of the hill is 45 degrees, I was able to work out the vertical distance from the x axis (at 4 metres) to where the water hits the hill using tan 45 degrees = opposite / adjacent = opposite / 4 metres, so the opposite is also equal to 4 metres. Therefore the total vertical height travelled by the water is 1 metre (the height as far as the x axis) plus 4 metres = 5 metres.
I then plugged this value in to the other standard equation for a projectile to calculate the total time that the water spends in the air (from leaving the hole until hitting the top if the hill) and again, since v sine theta = 0, this equation reduces to: t = 1 / g times the square root of 2 times g times h, which gives 1 second.
I was then able to write the (x,y) co-ordinates of where the water hits the hill (4 metres, - 4 metres)
However, for the final problem, I think that I can calculate a value for the velocity of water where it hits the hill, but I am not sure how to write it in vector form.

[ebaines]

The vector form of the velocity means the horizontal and vertical components of velocity separated out:



where are unit vectors in the horizontal and vertical directions, and are the magnitudes of the horizontal and vertical velocities, respectively. The magnitude of the horizontal velocity is constant, and the magnitude of the vertical velocity component is

[sheila1983]

The vector form of the velocity means the the horizontal and vertical components of velocity separated out:



where are unit vectors in the horizontal and vertical directions, and are the magnitudes of the horizontal and vertical velocities, respectively. The magnitude of the horizontal velocity is constant, and the magnitude of the vertical velocity component is

OK, thanks for explaining the form of the vectors.
I am just wondering, should Vy be equal to -gt,
since one of the equations of motion is V = u + at, where u is the initial velocity in the y direction = 0 and a = -g?

[ebaines]

Yes, I'm afraid I really messed that up! What I wrote is the equationn for vertical displacement, not velocity, Doh! The vertical velocity is indeed V_y= -gt.