/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 You drill a small hole in the si... [FREE SOLUTION] | 91Ó°ÊÓ

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You drill a small hole in the side of a vertical cylindrical water tank that is standing on the ground with its top open to the air. (a) If the water level has a height \(H,\) at what height above the base should you drill the hole for the water to reach its greatest distance from the base of the cylinder when it hits the ground? (b) What is the greatest distance the water will reach?

Short Answer

Expert verified
(a) Drill the hole at \( h = \frac{H}{2} \). (b) Greatest distance is \( H \).

Step by step solution

01

Understand the Problem

We are dealing with projectile motion of water exiting a cylindrical tank. We need to determine the optimal height for drilling a hole such that the horizontal distance traveled by the water is maximized.
02

Apply Torricelli's Law

The speed of water exiting the hole is given by Torricelli's Law, which states that the speed \( v \) is \( v = \sqrt{2g(H-h)} \), where \( h \) is the height of the hole above the water’s base, \( H \) is the height of the water, and \( g \) is the acceleration due to gravity.
03

Describe Vertical Motion

The water will fall from the height \( h \) to the ground. The time \( t \) it takes to fall can be calculated using the equation \( h = \frac{1}{2}gt^2 \), solving for \( t \) gives \( t = \sqrt{\frac{2h}{g}} \).
04

Calculate Horizontal Distance

The horizontal distance \( d \) travelled by the water can be calculated by multiplying its speed by the time of flight. Thus, \( d = v \cdot t = \sqrt{2g(H-h)} \cdot \sqrt{\frac{2h}{g}} = 2\sqrt{h(H-h)} \).
05

Maximize the Horizontal Distance

To find the height \( h \) that maximizes \( d \), we can differentiate \( 2\sqrt{h(H-h)} \) with respect to \( h \), and set the derivative to zero. This gives: \( \frac{d}{dh} [2\sqrt{h(H-h)}] = \frac{H-2h}{\sqrt{h(H-h)}} = 0 \). Solving for \( h \) gives \( h = \frac{H}{2} \).
06

Find the Greatest Distance

Substitute \( h = \frac{H}{2} \) into the expression for \( d \). This leads to \( d = 2\sqrt{\frac{H}{2}(H-\frac{H}{2})} = 2\sqrt{\frac{H}{2} \cdot \frac{H}{2}} = \frac{H}{2} \cdot 4 = H \). Thus, the greatest distance the water can reach is \( H \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Torricelli's Law
Torricelli's Law is a fundamental principle in fluid dynamics that helps us understand how liquids behave when exiting a container. Named after Evangelista Torricelli, a 17th-century Italian physicist, the law states that the speed (\( v \)) of a fluid flowing out of an orifice is tantamount to the speed that a freely falling object would have from the same height as the fluid column driving the flow.
In practical terms, if you have a tank filled with water to a height (\( H \)), and you drill a small hole at a height (\( h \)) above the tank's base, Torricelli's Law posits that the water will exit the hole with a speed of:
  • \( v = \sqrt{2g(H-h)} \)
Here, \( g \) is the acceleration due to gravity, approximately \( 9.81\, \text{m/s}^2 \).
This equation functions under the assumption that the size of the hole is small compared to the entire surface area of the fluid, and that the viscous forces and air resistance can be ignored. Understanding this concept provides crucial insight into optimizing the factors affecting how far and how fast water (or any liquid) can exit a reservoir.
Optimal Height Calculation
Determining the optimal height to drill a hole in a tank to achieve the maximum horizontal distance requires understanding both fluid and projectile physics.
In our problem involving a cylindrical water tank, we need the water stream to travel as far as possible horizontally. Using the insights from projectile motion, we realize that the height (\( h \)) at which we drill the hole plays a crucial role.
To ensure that the stream covers the greatest distance, we find that this optimal height is exactly half the water column's height, (\( H \)). The math behind it involves maximizing the horizontal distance formula:
  • \( d = 2\sqrt{h(H-h)} \)
Taking the derivative of \( d \) with respect to \( h \) and setting it equal to zero unveils that the drilling height should be:
  • \( h = \frac{H}{2} \)
At this height, the water divides its initial potential energy equally between the speed at which it exits the hole and the time it takes to hit the ground, creating the perfect conditions to maximize range.
Horizontal Distance Maximization
Maximizing horizontal distance in situations involving projectile motion can be a thrilling yet complex task. It essentially concerns identifying the correct environmental and physical configurations to extend the reach of the water stream.
In our water tank scenario, after using Torricelli's Law to find the exit speed and determining the optimal height of the hole, we employ the principles of projectile motion to focus on maximizing the distance.
The formula governing horizontal distance (\( d \)) is:
  • \( d = 2\sqrt{h(H-h)} \)
The goal is to achieve the highest value for \( d \). By choosing the hole height as half of the water column's height, both the vertical drop time and exit velocity are maximized within their constraints, leading to the effective horizontal movement.
When substituting \( h = \frac{H}{2} \) into the equation, the result leads to the maximum distance:
  • \( d = H \)
This means the water stream will travel a distance equal to the height of the water column, offering a practical illustration of how physics can predictably and repeatably govern real-world scenarios.

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Most popular questions from this chapter

Lift on an Airplane. Air streams horizontally past a small airplane's wings such that the speed is 70.0 \(\mathrm{m} / \mathrm{s}\) over the top surface and 60.0 \(\mathrm{m} / \mathrm{s}\) past the bottom surface. If the plane has a wing area of 16.2 \(\mathrm{m}^{2}\) on the top and on the bottom, what is the net vertical force that the air exerts on the airplane? The density of the air is 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) .

A cylindrical bucket, open at the top, is 25.0 \(\mathrm{cm}\) high and 10.0 \(\mathrm{cm}\) in diameter. A circular hole with a cross-sectional area 1.50 \(\mathrm{cm}^{2}\) is cutin the center of the bottom of the bucket. Water flows into the bucket from a tube above it at the rate of \(2.40 \times 10^{-4} \mathrm{m}^{3} / \mathrm{s}\) . How high will the water in the bucket rise?

Black Smokers. Black smokers are hot volcanic vents that emit smoke deep in the ocean floor. Many of them teem with exotic creatures, and some biologists think that life on earth may have begun around such vents. The vents range in depth from about 1500 \(\mathrm{m}\) to 3200 \(\mathrm{m}\) below the surface. What is the gauge pressure at a \(3200-\mathrm{m}\) deep vent, assuming the density of water does not vary? Express your answer in pascals and atmospheres.

In intravenous feeding, a needle is inserted in a vein in the patient's arm and a tube leads from the needle to a reservoir of fluid (density 1050 \(\mathrm{kg} / \mathrm{m}^{3} )\) located at height \(h\) above the arm. The top of the reservoir is open to the air. If the gauge pressure inside the vein is 5980 \(\mathrm{Pa}\) , what is the minimum value of \(h\) that allows fluid to enter the vein? Assume the needle diameter is large enough that you can ignore the viscosity (see Section 14.6 ) of the fluid.

A vertical cylindrical tank of cross-sectional area \(A_{1}\) is open to the air at the top and contains water to a depth \(h_{0}\) . A worker accidentally pokes a hole of area \(A_{2}\) in the bottom of the tank. (a) Derive an equation for the depth \(h\) of the water as a function of time \(t\) after the hole is poked. (b) How long after the hole is made does it take for the tank to empty out?

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