A water tank of height h has a small hole at height y. The wat...

Question

A water tank of height h has a small hole at height y. The water is replenished to keep h from changing. The water squirting from the hole has range 𝓍. The range approaches zero as y → 0 because the water squirts right onto the ground. The range also approaches zero as y → h because the horizontal velocity becomes zero. Thus there must be some height y between 0 and h for which the range is a maximum. Find an algebraic expression for the flow speed v with which the water exits the hole at height y.

Accepted Answer

Hey, everyone. Let's go through this practice problem. A cubicle water reservoir of side L is located on the roof of a building. The reservoir has developed a leak at a distance H from the roof's ground water is leaking from a small opening. The horizontal distance between the opening and the position where the water hits the ground is D, the water in the reservoir is constantly replenished by a pump to maintain a constant level at a distance of L from the roof's ground note that if the leak were at the bottom of the reservoir, the horizontal distance D would approach zero. The situation would be the same if the leak were at the top of the reservoir, the horizontal distance D would approach zero, determine the expression for the slow speed V at which the water exits the opening. And we have four multiple choice options to choose from. Option A, the square root of two G multiplied by L minus H. Option B the square root of two GL option C, the square root of two G multiplied by L minus D and option D, the square root of two G multiplied by D minus H So the first thing I'm gonna do before doing anything else is draw a bit of a diagram just so we're clear on what's happening in this problem. So it's a cubicle reservoir with a side length of L. So that's a, so I'm just gonna draw one face of it. It's cubicle. So the width of the reservoir, L is the same as the height of the reservoir again. L and that some height H there is a hole out of which water is flowing. And there's a horizontal distance from where the water lands of little D I better make a note in the diagram. The height from the bottom to where the hole is is H. Now, the problem is asking us to find the slow speed V. So the speed out of which the water flows through the hole, that means we're gonna want to find a relationship between that flow speed and other variables related to the way this tank is flowing. And one way we can do this is to use the Bernoulli equation which gives the relationship between a lot of different variables with problems like these with situations like this where we have flow happening. And the Bernoulli equation states that the pressure of the fluid plus one half the fluid's density multiplied by a flow speed squared plus the density multiplied the gravitational acceleration H multiplied by the vertical position of the point we're looking at is constant everywhere throughout the flow, we can pick two different points that are convenient for us in this tank of water and compare them to each other using the Bernoulli equation and use that to solve for the speed of the water through that hole. So I'm going to define two different points using the subscripts one and two. I'm going to say I'm gonna use the subscript two to refer to variables related to the point where the water is flowing out the hole. And I'm going to use the subscript one to refer to variables affecting the surface of the tank. I'm defining the two subscripts this way because both of those points have variables that are given to us in the problem to make it more clear what I mean. Let's go through and start writing out the Bernoulli equation for this problem. So piece of one, the pressure of one plus one half row, the sub one squared plus row G and then the vertical position for uh 0.1 is just the total height of the tank which is given as L and this is equal to P sub two plus one half row, the sub two squared plus row G. And then the vertical position of where the hole is, which is given to us in the problem as h the first thing we can do is notice the fact that since both of the points we've defined are exposed to the atmosphere, that means that P one and P two are the same because the pressure acting on both cases is the atmospheric pressure. So if P sub one and P sub two are equal, then we can cancel them out of the equation. Also notice that all of the remaining terms that are left to us include the density row, which means that they can all cancel out as well. Finally recall from the problem that we're being told that the water is being pumped into the tank so that a constant level is held of the water. This means that at 0.1 there is a net flow speed of zero because the water at that point is staying the same. So and that means that the sub one goes to zero. So the entire one half row V one squared term disappears. So all that we're left with is an equation saying that GL is equal to one half V sub two squared plus GH. Let's now begin solving this equation for V sub two. First, let's get both of the non no. First, let's get the term containing V sub two on its own by subtracting away everything else. So one half V sub two squared is equal to GL minus GH. And we can simplify this a bit by factoring out the G on the right hand side of the equation. So that's equal to G multiplied by goddamn it. So let's start by getting the term containing V sub two on its own, we can do that by subtracting from both sides of the equation, the GH term. So we find an equation saying that one half V sub two squared is equal to GL minus GH, we can simplify this further by factoring out the G so G multiplied by L minus H. And then I'm gonna multiply both sides of the equation by two to get rid of this one half. So V sub two squared is equal to two G L minus H. And then lastly, we can get the V sub two completely on its own by taking the square root of both sides of the equation. So V sub two is equal to the square root of two GL minus H. And now we found an equation for V sub two which gives us the flow speed of the water out from the hole at 0.2. So this equation here is our final answer for the problem. And if we look at our options, we can see that one of the options we're given in the problem is exactly this option A says that the square root of two G multiplied by L minus H. So part A is our answer to this problem. And that's it for this problem. I hope this video helped you out if it did. And you'd like more practice. Please check out some of our other videos which will hopefully give you more experience with these types of problems, but that's all for now. I hope you all have a lovely day.

Key Concepts

Bernoulli's Principle

Torricelli's Law

Projectile Motion

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