/*! 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 49 In fighting forest fires, airpla... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 49

In fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. For practice, a pilot drops 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 has 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.

Short Answer

Expert verified
The pilot should release the canister approximately 273.92 m from the target to hit it accurately.

Step by step solution

01

Calculate the Fall Time

The first step is to calculate how long the canister takes to fall. This can be done using the formula for the time it takes an object to fall from a certain height under gravity: \( t = \sqrt{\frac{2h}{g}} \), where \( h = 90.0 \, \mathrm{m} \) is the altitude of the plane and \( g = 9.81 \, \mathrm{m/s^2} \) is the acceleration due to gravity.
02

Compute the Fall Time

Substitute \( h = 90.0 \, \mathrm{m} \) and \( g = 9.81 \, \mathrm{m/s^2} \) into the formula to find the time : \( t = \sqrt{\frac{2 * 90.0 \, \mathrm{m}}{9.81 \, \mathrm{m/s^2}}} \). The result is approximately \( t = 4.28 \, \mathrm{s} \).
03

Calculate the Horizontal Distance

Now, use the speed of the plane and the time it takes for the canister to fall to calculate the horizontal distance from the target where the canister should be dropped. Since the horizontal distance is the product of speed and time, use the equation \( d = vt \), where \( v = 64.0 \, \mathrm{m/s} \) is the speed of the plane and \( t = 4.28 \, \mathrm{s} \) is the time it takes for the canister to fall. So, \( d = (64.0 \, \mathrm{m/s}) * (4.28 \, \mathrm{s}) \).

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.

gravity
When discussing projectile motion, gravity plays a crucial role. It is the force that pulls objects toward the Earth's center, causing them to fall when dropped or thrown. In the given problem, we are dealing with a canister dropped from an airplane. Since gravity is pulling it towards the ground, it becomes a vital factor in calculating the time it takes for the canister to hit the ground from a height of 90 meters.
To understand the effects of gravity, we use the formula for an object's fall time: \( t = \sqrt{\frac{2h}{g}} \), where \( h \) is the height (90 meters in this case), and \( g \) is the acceleration due to gravity, which is approximately 9.81 m/s虏 on Earth. This formula helps to determine how long the canister will take to fall, which is essential for deciding when to release it from the airplane. Without considering gravity, the canister's path to the ground would be miscalculated.
horizontal distance
In projectile motion, horizontal distance refers to how far an object travels along the horizontal plane before hitting the ground. For our canister dropped from an airplane, calculating this is essential to ensure it hits the target on the ground. The key to determining horizontal distance is understanding that, in the absence of air resistance, horizontal velocity remains constant.
The formula we use here is \( d = vt \), where \( d \) is the horizontal distance, \( v \) is the horizontal velocity (64 m/s for the airplane), and \( t \) is the time calculated previously (4.28 seconds). By substituting these values, you find the horizontal distance the canister travels before touching the ground. This precise calculation allows the pilot to time the release optimally.
vertical motion
Vertical motion involves the movement of an object as it falls under the influence of gravity, excluding any initial vertical velocity, since our problem assumes a horizontal drop. This means that the only force acting vertically on the canister is gravity, causing it to accelerate downward.
In physics, this vertical fall is crucial for calculating when and how objects will hit the ground. For the canister, knowing the time it takes to fall (calculated using gravitational formulas) informs both when the object should be released and how gravity impacts its descent. Studying vertical motion helps to understand the downward trajectory and ensures our canister lands precisely at the target location. Get familiar with these vertical concepts to master relating them to real-world physics problems like this.

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

A rhinoceros is at the origin of coordinates at time \(t_{1}=0 .\) For the time interval from \(t_{1}=0\) to \(t_{2}=12.0 \mathrm{~s},\) the rhino's average velocity has \(x\) -component \(-3.8 \mathrm{~m} / \mathrm{s}\) and \(y\) -component \(4.9 \mathrm{~m} / \mathrm{s}\). At time \(t_{2}=12.0 \mathrm{~s},\) (a) what are the \(x\) - and \(y\) -coordinates of the rhino? (b) How far is the rhino from the origin?

A small object is projected from level ground with an initial velocity of magnitude \(16.0 \mathrm{~m} / \mathrm{s}\) and directed at an angle of \(60.0^{\circ}\) above the horizontal. (a) What is the horizontal displacement of the object when it is at its maximum height? How does your result compare to the horizontal range \(R\) of the object? (b) What is the vertical displacement of the object when its horizontal displacement is \(80.0 \%\) of its horizontal range \(R ?\) How does your result compare to the maximum height \(h_{\max }\) reached by the object? (c) For when the object has horizontal displacement \(x-x_{0}=\alpha R,\) where \(\alpha\) is a positive constant, derive an expression (in terms of \(\alpha\) ) for \(\left(y-y_{0}\right) / h_{\max }\). Your result should not depend on the initial velocity or the angle of projection. Show that your expression gives the correct result when \(\alpha=0.80,\) as is the case in part (b). Also show that your expression gives the correct result for \(\alpha=0, \alpha=0.50\), and \(\alpha=1.0\)

A dog running in an open field has components of velocity \(v_{x}=2.6 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=-1.8 \mathrm{~m} / \mathrm{s}\) at \(t_{1}=10.0 \mathrm{~s} .\) For the time interval from \(t_{1}=10.0 \mathrm{~s}\) to \(t_{2}=20.0 \mathrm{~s},\) the average acceleration of the dog has magnitude \(0.45 \mathrm{~m} / \mathrm{s}^{2}\) and direction \(31.0^{\circ}\) measured from the \(+x\) -axis toward the \(+y\) -axis. At \(t_{2}=20.0 \mathrm{~s},\) (a) what are the \(x\) - and \(y\) -components of the dog's velocity? (b) What are the magnitude and direction of the dog's velocity? (c) Sketch the velocity vectors at \(t_{1}\) and \(t_{2}\). How do these two vectors differ?

A jet plane is flying at a constant altitude. At time \(t_{1}=0,\) it has components of velocity \(v_{x}=90 \mathrm{~m} / \mathrm{s}, v_{y}=110 \mathrm{~m} / \mathrm{s} .\) At time \(t_{2}=30.0 \mathrm{~s}\) the components are \(v_{x}=-170 \mathrm{~m} / \mathrm{s}, v_{y}=40 \mathrm{~m} / \mathrm{s}\). (a) Sketch the velocity vectors at \(t_{1}\) and \(t_{2} .\) How do these two vectors differ? For this time interval calculate (b) the components of the average acceleration, and (c) the magnitude and direction of the average acceleration.

Canada geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at speeds up to about \(100 \mathrm{~km} / \mathrm{h}\). If one goose is flying at \(100 \mathrm{~km} / \mathrm{h}\) relative to the air but a \(40 \mathrm{~km} / \mathrm{h}\) wind is blowing from west to east, (a) at what angle relative to the north-south direction should this bird head to travel directly southward relative to the ground? (b) How long will it take the goose to cover a ground distance of \(500 \mathrm{~km}\) from north to south? (Note: Even on cloudy nights, many birds can navigate by using the earth's magnetic field to fix the northsouth direction.)
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Famous Physicists

Read Explanation

Space Physics

Read Explanation

Thermodynamics

Read Explanation

Electrostatics

Read Explanation

Energy Physics

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.