Video Transcript
A tall water container has small holes in its side at different heights above the ground, as shown in the diagram. Water leaks from the holes in the container, and only the leak from hole 𝐴 is shown. The leaks from the holes travel different distances horizontally from the water container. The volume of water leaking from the container is assumed to be small so that the decrease in the height of the water column due to the leaking water is negligible. From which hole would the water travel farthest horizontally from the container?
Here, we see our leaking water container with four holes 𝐴, 𝐵, 𝐶, and 𝐷 in its side. Water leaks from all four holes, and those four leaks travel different distances horizontally before they reach the ground. We want to know from which of these four holes would water travel farthest horizontally. We can call the horizontal distance traveled by each stream of water its range. If we think about what makes water leak out of this container, it has to do with the pressure of the water in the column. For example, if we consider the water that we know to be coming out of hole 𝐷, all this fluid here is creating pressure to force water out of that hole.
Clearing some space on screen to work, we can recall that the pressure created by a fluid of density 𝜌 is equal to that density multiplied by the acceleration due to gravity 𝑔 times the depth 𝑑 of the fluid. If we say that our column of water has an overall height capital 𝐻 and this height, we’ll assume, doesn’t change even as water slowly leaks out of these small holes, and if we assume further that the distance from ground level to a given hole in the container is lowercase ℎ, then in that case the depth 𝑑 of a given hole is equal to capital 𝐻 minus lowercase ℎ.
The application of Pascal’s principle to our scenario then is to say that the pressure that forces water out of any given hole in the container is equal to the density of water 𝜌 times the acceleration due to gravity 𝑔 multiplied by the depth of that hole capital 𝐻 minus lowercase ℎ. We can get an idea of the effect of this pressure by taking an up-close view at one of the holes in the wall of our container. Here is the container wall, and here is the small opening in that wall that lets water leak out.
Up to this point, we can assume that the water to the left of this line is static, not moving. But then, thanks to this pressure 𝑃 created by all the fluid above this hole in the column of water, not only is a downward pressure produced at this height, but actually an equal pressure in all directions is generated due to the pressure 𝑃. If this were not the case, if the pressure, say, were greater in one direction than in another, then that would make the fluid at this point move. But we’re assuming that this fluid is stationary.
When this pressure acts on a particle of water that’s located here though, that particle does not stay still. Rather, it accelerates to the right through the wall of the container. We’ll assume that when this particle of water reaches the outer wall of the container, its horizontal acceleration stops and it enters freefall. That then creates the stream of water that we would see falling out of the side of the container.
Now, the greater the pressure 𝑃, that is, the greater the depth of a given hole, the stronger the push on a particle of water will be to accelerate it. From that perspective, we can say that the lower down in the container wall a given hole is, the greater the exit speed of that water will be from the container. So, for example, water coming out of hole 𝐴, which doesn’t have very much water creating pressure pushing the water through that hole, will have a relatively small exit speed, while water from hole 𝐷 will have a relatively greater exit speed because there’s more pressure pushing that water through.
This doesn’t necessarily mean though that the water leaking out of hole 𝐷 will travel farther horizontally, that is, have a greater range than water from any of the other holes. Even though water from hole 𝐷 will be traveling faster horizontally than water leaking out of any of the other holes, it will also have less time to move horizontally before it reaches ground level. That is, the time it takes to fall down will be less time than the water from 𝐶, 𝐵, or 𝐴 has to fall.
Here’s what we can write. The range of a stream of water coming out of a given hole is equal to the exit velocity of that water, we’ve called it 𝑣 sub h because it’s a horizontal velocity, multiplied by the time it takes water leaking from that hole to fall to the ground. Note by the way that this equation is an application of the relationship where the average speed 𝑣 of an object is equal to the distance that object travels divided by the time taken to travel that distance. Here, the horizontal velocity of an exiting stream of water is constant over its descent. And so if we multiply that velocity by the time it takes water to fall to the ground from that given hole, then the product is equal to the range of the water, the horizontal distance traveled.
So far, we’ve seen that the greater the depth of a hole in our container’s wall, the greater the horizontal exit velocity of water from that hole will be. On the other hand, water from this relatively farther down hole will have less time to fall before it reaches the ground. Water from the hole that has the greatest range will give us the largest result from this function of two competing variables.
Let’s look more closely at 𝑣 sub h and 𝑡 sub fall then, starting with the horizontal exit velocity. We’ve said that for a particle of water that’s just on the border of being stationary and being accelerated into motion, that particle’s motion is due to the pressure created by the fluid around it. Along with knowing that that pressure is equal to the density of water times the acceleration due to gravity times the quantity capital 𝐻 minus lowercase ℎ, we can recall from Pascal that pressure in general is equal to a force spread out over an area. Or rearranging this slightly, force is equal to pressure times area.
In our case, that area is the cross-sectional area of our column of fluid. If we call that area capital 𝐴 and we know that it’s constant all throughout the height of our container, then we can say that the force that acts on our particle of water 𝐹 is equal to the pressure 𝜌 times 𝑔 times capital 𝐻 minus lowercase ℎ all multiplied by the cross-sectional area of our column of fluid.
As a next step, let’s remember Newton’s second law of motion that the net force on an object equals that object’s mass times its acceleration. Dividing both sides of this equation by mass 𝑚, we find that acceleration equals force divided by mass. So then, the acceleration of a particle of water just as it’s beginning to move is equal to the density of water times the acceleration due to gravity multiplied by capital 𝐻 minus lowercase ℎ all times the cross-sectional area 𝐴 divided by the mass of that particle of water 𝑚.
Here’s something important to notice about this acceleration. At a given height in our container, that is, for a given value of lowercase ℎ, this acceleration 𝑎 is constant. That means that as our particles accelerated through this opening in the container wall, its motion can be described by what are sometimes called the kinematic equations of motion. These equations after all apply to the motion of particles whose acceleration is constant, like our water particle here.
In general, there are four kinematic equations of motion. But here we’ll just consider one of them. This equation says that for an object with constant acceleration, the final velocity of that object squared equals its initial velocity squared plus two times its acceleration times its displacement 𝑠. When we apply this equation to our scenario, we’re doing it so we can solve for the horizontal velocity of our particle of water right here as it leaves the container wall. We’ve called this velocity 𝑣 sub h. So here we’re solving for 𝑣 sub h squared.
When our particle of water first begins accelerating, its initial speed and its initial velocity is zero. Therefore, 𝑣 sub h squared equals zero squared plus two times 𝑎 times 𝑠. Here, the acceleration of our water particle is given by this expression, while the displacement 𝑠 is the thickness of the wall of our container. So then, 𝑣 sub h squared equals two times 𝑎 times 𝑠. And when we substitute in our value for the acceleration 𝑎, we get this overall expression for 𝑣 sub h squared. In our range equation though, we see that we don’t have 𝑣 sub h squared, but just 𝑣 sub h. What we’ll do then is take the square root of both sides of this equation so that on the left-hand side the square of 𝑣 sub h and the square root cancel one another out.
We now have an expression for 𝑣 sub h, though it looks pretty complicated. Let’s remember though that most of the values in this expression are constants. All we’re really concerned about keeping track of, as far as its effect on 𝑣 sub h, is the value in the parentheses, capital 𝐻 minus lowercase ℎ. In fact, clearing some space on screen, we can even rewrite 𝑣 sub h to emphasize this fact. Here, we have our first square root, where all of the quantities which don’t change are included. This is multiplied by the second square root, where we do have a quantity that can change, lowercase ℎ. This then is our expression for 𝑣 sub h.
Now, let’s find one for 𝑡 sub fall, the time it takes water leaking out of a given hole to fall all the way to the ground. As water falls out of a given hole, its acceleration is constant. It equals the acceleration due to gravity 𝑔. Therefore, we can describe this falling water using a kinematic equation of motion once more. This time, the equation we use says that the displacement of an object equals that object’s initial velocity times the time elapsed plus one-half that object’s acceleration 𝑎 times the time elapsed squared. In our application, that displacement is the vertical displacement of the water as it falls to the ground. In other words, it’s the height of the hole in the container wall, lowercase ℎ.
As we’ve seen, when water first leaks out of a hole, it only has horizontal velocity and no vertical velocity. Therefore, 𝑣 sub i, the velocity in this case in the vertical direction of our water, is zero. That doesn’t stay true for long though due to the acceleration caused by gravity 𝑔. In our application, the time elapsed 𝑡 is equal to 𝑡 sub fall, the time it takes water to fall to the ground from a hole. This then is how we would write out this kinematic equation in our case. ℎ is equal to one-half 𝑔 times 𝑡 sub fall squared. If we multiply both sides of this equation by two divided by 𝑔, then two multiplied by one-half on the right is equal to one and 𝑔 divided by 𝑔 is equal to one. Therefore, 𝑡 fall squared is equal to two times ℎ divided by 𝑔. If we take the square root of both sides, then the square on 𝑡 sub fall and the square root cancel out. And now we have a simplified expression for 𝑡 sub fall.
Now that we have expressions for both 𝑡 sub fall and 𝑣 sub h, let’s see how they combine together to give an overall range 𝑅. Considering the product of these two values, note that we have a square root of 𝑔 and a square root of one over 𝑔. So the product of these two will equal one. We notice also that we have a square root of two times the square root of two. That means we can factor out a two from under the square root sign. In order to see how range is maximized, really what we want to focus on here is the product of these two square roots. In fact, we can rewrite these so that they’re under one square root so that we have all of these constants here multiplied by the square root of the quantity capital 𝐻 minus lowercase ℎ times lowercase ℎ.
Let’s recall that lowercase ℎ is the height above ground level at which a given hole exists. In solving for which of our four holes leaks water with the greatest horizontal distance, we’re essentially varying lowercase ℎ and seeing for which value of lowercase ℎ this expression is maximized. Just to get a sense for how this works, let’s imagine substituting in a couple of different values for lowercase ℎ. Let’s say that lowercase ℎ equals zero. That is, we have a hole that is at the base of our column of water. In that case, this value right here is zero. And therefore the value under the square root is zero, and so the total range overall is also zero. Since the range of water from that hole is zero, that can’t be our maximum horizontal distance.
Now, let’s try a different extreme. Let’s say the hole is positioned at the top of the column. In other words, lowercase ℎ is equal to capital 𝐻. If that’s so, then this lowercase ℎ would be equal to capital 𝐻. And therefore we would have capital 𝐻 minus capital 𝐻, or zero. So, if lowercase ℎ were this maximum value of capital 𝐻, then the range once again would be zero. That’s because there’ll be no pressure to push water horizontally out of a hole in the container wall. The value of lowercase ℎ that maximizes this expression will be somewhere between zero and capital 𝐻. And in fact if we let lowercase ℎ equal exactly half of capital 𝐻, then that leads to a maximum value of this expression under the square root sign. Specifically, it equals capital 𝐻 minus capital 𝐻 divided by two all multiplied by capital 𝐻 over two. That product is equal to capital 𝐻 squared over four.
In this question, we don’t need to solve for that maximum range value 𝑅. We just want to know which of our four holes will leak water that has the greatest range. Based on our analysis, the hole that has a height closest to capital 𝐻 over two, half the overall height of the water column, will leak water that has the greatest horizontal range. Looking at the four holes in our container, we see that hole 𝐷 is very close, if not located at, the midpoint of this water column. Therefore, we’ll say that water leaked through this hole has the greatest range. Of the water that leaks through holes 𝐴, 𝐵, 𝐶, and 𝐷, the water that leaks through hole 𝐷 will travel the greatest distance horizontally before it reaches the ground.