# Max Height Attained by Water Stream in P9.81 Tank

• parwana
In summary, if you open the valve on the water tank shown in Figure P9.81, the water stream will reach its maximum height at h = 8.0 m, L = 1.50 m, and = 30°.
parwana

## Homework Statement

Figure P9.81 shows a water tank with a valve at the bottom. If this valve is opened, what is the maximum height attained by the water stream coming out of the right side of the tank? Assume that h = 8.0 m, L = 1.50 m, and = 30° and that the cross-sectional area at A is very large compared with that at B.

## Homework Equations

Bernouli's equation
P1 + 1/2(density)(v1)^2 + density(g)(y)= P2 + 1/2(density)(v2)^2 + density(g)(y)

## The Attempt at a Solution

0+0+density(g)(y)= 0+1/2(density)(v2)^2 + density(g)(Lsin30)

I tried solving for v2

I am stuck now at this step

used that v2 as v initial to find y by

vfinal^2= vinitial^2-2g(y final-y initial)

I am not sure but would v initial be the v2 I found? And is v final 0? Also what is y final and what is y initial?

HELP

Why do you think P1 and P2 are zero?. The water is being launched at an angle. Only the vertical component enters into the height calculation.

so the vertical component, which is Lsin30, is that the initial height? I am confused, I tried this again and again but I can't egt it right. And the gauge pressure is 0 since the air is touching the water, air's pressure is 1 atm

parwana said:
so the vertical component, which is Lsin30, is that the initial height? I am confused, I tried this again and again but I can't egt it right. And the gauge pressure is 0 since the air is touching the water, air's pressure is 1 atm

I assume they want the maximum height measured from the same place as the 8m height, but maybe they want it measured from where it leaves the spout. The second is easier, but the first can be obtained by just adding Lsinθ. You are OK on the pressure. I was misinterpreting. The height changes from 8m to Lsinθ to the height relative to the end of the spout. Using that (as I think you have done) you can solve for v at the exit of the spout. The vertical component of that velocity is v*sinθ. You can compute the height relative to the end of the spout by finding the point where the vertical velocity is zero. You can either do that by computing the time it takes to reach that point and plugging that into the y(t) equation, or you can use the equation that relates the change in the velocity squared to acceration and distance.

Last edited:
Hey, I solved it myself and get 1.8125 metres (I presume you already know the answer and need the method). Your method is flawed.

The Bernoulli equation tells us that the mechanical energy per unit mass, let's call it E is:

E = gz + v2/2 + (p-p0)/rho.

E = gravitational potential energy + kinetic energy + pressure energy.

When you open the valve you just say that the mechanical energy at A is the same as that at B. Because the pressure energy is the same at these points they cancel from both sides. Furthermore, because the cross-sectional at A is much larger than that at B you can ignore the kinetic energy at A. You end up with a fairly simple equation that you have to solve for velocity. Once you've got that velocity you need to find how high it gets which will involve a little projectile mechanics, if I could do it I'm sure you can.

thank u so much guys, I understand the method better now

## 1. What is the "Max Height Attained by Water Stream in P9.81 Tank" phenomenon?

The "Max Height Attained by Water Stream in P9.81 Tank" refers to the maximum height that a water stream can reach when released from a tank with a gravitational pull of 9.81 meters per second squared.

## 2. How is the max height attained by the water stream calculated?

The maximum height is calculated using the equation h = (v^2)/(2g), where h is the height, v is the initial velocity of the water stream, and g is the gravitational pull.

## 3. What factors affect the max height attained by the water stream?

The factors that affect the maximum height attained by the water stream include the initial velocity of the water, the diameter of the tank, the air pressure inside the tank, and the angle at which the water is released.

## 4. Why is the gravitational pull in the tank specified as P9.81?

P9.81 refers to the standard acceleration due to gravity, which is 9.81 meters per second squared. This value is used as the gravitational pull in the tank to account for the Earth's natural gravitational force.

## 5. How is the "Max Height Attained by Water Stream in P9.81 Tank" phenomenon useful in scientific research?

The "Max Height Attained by Water Stream in P9.81 Tank" phenomenon is useful in understanding the principles of projectile motion and gravity. It can also be used to study the effects of different factors, such as air pressure and angle of release, on the maximum height reached by a water stream. This information can be applied in various fields, such as fluid dynamics, engineering, and physics.

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