/*! 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 60 A water hose is used to fill a l... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 60

A water hose is used to fill a large cylindrical storage tank of diameter \(D\) and height 2\(D .\) The hose shoots the water at \(45^{\circ}\) the horizontal from the same level as the base of the tank and is a distance 6\(D\) away (Fig. P3.60). For what range of launch speeds \(\left(v_{0}\right)\) will the water enter the tank? Ignore air resistance, and express your answer in terms of \(D\) and \(g .\)

Short Answer

Expert verified
The hose launch speed must be between \(\sqrt{\frac{gD}{3.25}}\) and \(\sqrt{4.8075 g D}\).

Step by step solution

01

Calculate the horizontal distance to the tank

The water from the hose travels horizontally a distance of 6\(D\) plus the radius of the tank \(0.5D\), since the water must reach the side of the tank. Therefore, the total horizontal distance the water must cover for entering the tank is\[6D + 0.5D = 6.5D.\]
02

Calculate the time of flight

The projectile's time of flight \(t\) is related to the horizontal distance \(d\) as follows:\[ d = v_0 \cos(45^\circ) \cdot t = 6.5D \]Thus, expressing \(t\) in terms of \(v_0\):\[ t = \frac{6.5D}{v_0 \cdot \cos(45^\circ)} = \frac{6.5D}{\frac{v_0}{\sqrt{2}}} = \frac{6.5D\sqrt{2}}{v_0}\]
03

Determine vertical climb condition

The water must rise to the height that includes the top of the tank. This vertical height \(H\) is precisely the height of the tank, \(2D\). The vertical position of the water can be described as:\[ y = v_0 \sin(45^\circ) t - \frac{1}{2} g t^2 \]Substituting \(t\) from the previous step into \(y = 2D\):\[ v_0 \frac{1}{\sqrt{2}} \cdot \frac{6.5D\sqrt{2}}{v_0} - \frac{1}{2} g \left(\frac{6.5D\sqrt{2}}{v_0}\right)^2 = 2D \]
04

Simplify and solve for initial speed range

Simplify the expression:\[ 6.5D - \frac{g (6.5D\sqrt{2})^2}{2 v_0^2} = 2D \]Setting this equal and solving for \(v_0\) over its quadratic form, and solving gives:\[ v_0^2 > \frac{g \cdot 2D}{6.5} \text{ and } v_0^2 < \frac{g \cdot 2D}{{(6.5/2)}^2} \approx \frac{gD}{0.52} \]Thus, the ranging speed is:\[ \sqrt{\frac{g \cdot D}{3.25}} < v_0 < \sqrt{\frac{19.23g \cdot D}{4}} = \sqrt{4.8075gD} \]
05

Compose the final answer

The water will enter the tank if the launch speed is in the calculated range. Thus, the speed range is:\[ \sqrt{\frac{gD}{3.25}} < v_0 < \sqrt{4.8075 g D} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Cylindrical Tank
A cylindrical tank is a type of container with a circular base and straight sides. It is shaped like a cylinder, hence its name. These tanks are common in industries and homes for storing liquids like water, oil, or chemicals. Understanding its dimensions is crucial when dealing with projectile problems.

In our problem, the tank has a diameter of \(D\) and a height of \(2D\). This means if you were to look at the tank from above, you would see a circle with diameter \(D\). From the side, the height goes up to \(2D\). Visually picturing the tank can help when calculating how a projectile might interact with it, such as when water from a hose needs to enter it. The tank's radius, which is half of the diameter, is helpful in calculations and is \(0.5D\).
Horizontal Distance Calculation
Calculating the horizontal distance a projectile travels is crucial in projectile motion problems. The water in our problem needs to reach a tank from a hose placed at a distance. This problem involves calculating this distance precisely.

The hose is positioned at a distance of \(6D\) from the base of the cylindrical tank. To ensure the water enters the tank, you need to consider the additional distance to reach the tank's side. This additional distance is half the tank's diameter, adding \(0.5D\) to the initial \(6D\). Thus, the total horizontal distance the water must cover is \(6.5D\).
  • Initial set up distance: \(6D\)
  • Additional radius distance: \(0.5D\)
  • Total horizontal distance: \(6.5D\)
Vertical Motion Analysis
In projectile motion, vertical analysis is as important as horizontal. For a projectile like water from a hose, you analyze how high it rises when in motion.

The projectile needs to reach a certain height to ensure it enters the tank. The tank in question is \(2D\) high. The projectile鈥檚 vertical motion can be described using its initial vertical velocity and the influence of gravity.

The equation for vertical position is given by:
  • Initial vertical velocity component: \(v_0 \sin(45^\circ)\)
  • Vertical displacement formula: \(y = v_0 \sin(45^\circ) t - \frac{1}{2} g t^2\)
Inserting the time from the horizontal equation and setting \(y\) to \(2D\) allows us to solve for the initial velocity \(v_0\).
Initial Speed Range
Determining the range of initial speeds that allow a projectile to successfully reach a target is key in projectile motion problems. For the water to enter the tank, the initial speed must be within a specific range.

Using the previous calculations, the problem derives an inequality to express the range. The projectile must have enough speed to reach over \(6.5D\) horizontally and climb the tank height of \(2D\). Solving the set equations provides the range of speeds for \(v_0\).

The valid speed range can be expressed as:
  • Minimum speed: \(\sqrt{\frac{gD}{3.25}}\)
  • Maximum speed: \(\sqrt{4.8075gD}\)
This range ensures the water trajectory intersects with the tank's opening, making it crucial to calculations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Win the Prize. In a carnival booth, you win a stuffed giraffe if you toss a quarter into a small dish. The dish is on a shelf above the point where the quarter leaves your hand and is a horizontal distance of 2.1 \(\mathrm{m}\) from this point (Fig. E3.19). If you toss the coin with a velocity of 6.4 \(\mathrm{m} / \mathrm{s}\) at an angle of \(60^{\circ}\) above the horizontal, the coin lands in the dish. You can ignore air resistance. (a) What is the height of the shelf above the point where the quarter leaves your hand? (b) What is the vertical component of the velocity of the quarter just before it lands in the dish?

A canoe has a velocity of 0.40 \(\mathrm{m} / \mathrm{s}\) southeast relative to the earth. The canoe is on a river that is flowing 0.50 \(\mathrm{m} / \mathrm{s}\) east rela- tive to the earth. Find the velocity (magnitude and direction) of the canoe relative to the river.

The radius of the earth's orbit around the sun (assumed to be circular) is \(1.50 \times 10^{8} \mathrm{km},\) and the earth travels around this orbit in 365 days. (a) What is the magnitude of the orbital velocity of the earth, in \(\mathrm{m} / \mathrm{s} ?\) (b) What is the radial acceleration of the earth toward the sun, in \(\mathrm{m} / \mathrm{s}^{2} ?\) (c) Repeat parts (a) and (b) for the motion of the planet Mercury (orbit radius = \(5.79 \times 10^{7} \mathrm{km}\) , orbital period \(=88.0\) days \() .\)

A rocket is fired at an angle from the top of a tower of height \(h_{0}=50.0 \mathrm{m} .\) Because of the design of the engines, its position coordinates are of the form \(x(t)=A+B t^{2}\) and \(y(t)=C+D t^{3},\) where \(A, B, C,\) and \(D\) are constants. Furthermore, the acceleration of the rocket 1.00 s after firing is \(\vec{a}=(4.00 \hat{\imath}+3.00 \hat{J}) \mathrm{m} / \mathrm{s}^{2} .\) Take the origin of coordinates to be at the base of the tower. (a) Find the constants \(A, B, C,\) and \(D,\) including their SI units. (b) At the instant after the rocket is fired, what are its acceleration vector and its velocity? (c) What are the \(x\) -and \(y\) -components of the rocket's velocity 10.0 s after it is fired, and how fast is it moving? (d) What is the position vector of the rocket 10.0 s after it is fired?

In fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. A pilot is practicing by dropping a canister of red dye, hoping to hit a target on the ground below. If the plane is flying in a horizontal path 90.0 \(\mathrm{m}\) above the ground and with a speed of \(64.0 \mathrm{m} / \mathrm{s}(143 \mathrm{mi} / \mathrm{h}),\) at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Measurements

Read Explanation

Electromagnetism

Read Explanation

Radiation

Read Explanation

Geometrical and Physical Optics

Read Explanation

Wave Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.