/*! 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 70 A plane flies at \(1.25\) times ... [FREE SOLUTION] | 91Ó°ÊÓ

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A plane flies at \(1.25\) times the speed of sound. Its sonic boom reaches a man on the ground \(1.00\) min after the plane passes directly overhead. What is the altitude of the plane? Assume the speed of sound to be \(330 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The calculation error occurred because the defined description for altitude didn't correlate: recheck plane versus 0 interaction specific or 60 provisional balance.

Step by step solution

01

Determine the Plane's Speed

The speed of sound is given as \( v = 330 \, \mathrm{m/s} \). Since the plane flies at \( 1.25 \) times the speed of sound, we calculate the plane's speed as \( v_{plane} = 1.25 \times 330 \, \mathrm{m/s} = 412.5 \, \mathrm{m/s} \).
02

Calculate the Distance Covered by the Plane

The sonic boom reaches the man on the ground 1 minute (or 60 seconds) after the plane passes overhead. The distance covered by the plane during this time can be found using speed \( = \) distance/time. Thus, the distance covered by the plane in that time is \( d_{plane} = v_{plane} \times t = 412.5 \, \mathrm{m/s} \times 60 \, \mathrm{s} = 24750 \, \mathrm{m} \).
03

Establish the Relationship Using the Geometry of the Situation

When the sonic boom reaches the man on the ground, a right triangle is formed where the altitude of the plane is one leg, the distance covered by the plane is the other leg, and the distance the sound travels is the hypotenuse. The sound travels for 60 seconds, so the distance is \( d_{sound} = v \times t = 330 \, \mathrm{m/s} \times 60 \, \mathrm{s} = 19800 \, \mathrm{m} \).
04

Use the Pythagorean Theorem to Solve for Altitude

In the right triangle with sides representing altitude \( h \), the plane's distance covered \( 24750 \, \mathrm{m} \), and the sound's distance \( 19800 \, \mathrm{m} \), we apply the Pythagorean theorem:\[19800^2 = 24750^2 + h^2\]Solving for \( h \):\[h^2 = 19800^2 - 24750^2\]\[h^2 = 392040000 - 612562500\]\[h^2 = -220522500\]Since this doesn't resolve correctly check the usage of the plane overtaking ahead or equal speeds matching angle distance.
05

Fix Calculation Adjustments (02 planes same height and distance reached by sound)

On fixing we use variables to determine if recalculation is necessary. Hypotenuse being sound covering both total removal of excess means sound does not cover with violation till reduced or equal in capacity analysis.

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Key Concepts

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

Speed of Sound
The speed of sound is a fundamental concept in physics, referring to the speed at which sound waves propagate through a medium. In this exercise, the speed of sound is provided as 330 meters per second (\( ext{m/s} \)). This value depends significantly on factors like air temperature, pressure, and humidity. However, for simplicity, the exercise assumes a standard speed of sound.
When a plane flies faster than the speed of sound, it is called supersonic flight. The plane in our example flies at 1.25 times the speed of sound, leading to a faster-than-sound scenario.
  • Supersonic flights: These occur when an aircraft travels faster than the speed of sound.
  • Sonic boom: A loud noise associated with shock waves created when an object travels through the air faster than the speed of sound.
This results in the characteristic sonic boom heard after the aircraft has already passed overhead. Understanding how the speed of sound functions is crucial when solving problems involving scenarios like this.
Pythagorean Theorem
The Pythagorean Theorem is an essential mathematical principle used to define the relationship between the sides of a right triangle. The theorem is expressed as:
\[ c^2 = a^2 + b^2 \]Where \( c \) is the hypotenuse, and \( a \) and \( b \) are the other two sides. In the plane's scenario, we consider the distance the plane covered as one side \( b \), the altitude of the plane as the other side \( a \), and the distance the sonic boom traveled as the hypotenuse \( c \).
  • Right Triangle Identification: The exercise forms a right triangle scenario with the altitude forming one side.
  • Applying theorem: Using the theorem helps us calculate the unknown side if the other two sides are known.
The initial attempt to solve it using the Pythagorean theorem led to incorrect results due to misinterpretation of triangle distances. Correct application requires understanding the relative positions and distances involved, ensuring the computed values align with given conditions.
Altitude Calculation
Calculating the altitude of a supersonic plane requires understanding the relationships between speed and distance in a two-dimensional plane model. Here, the known quantities include the speed of sound, the time for the sound to reach the observer, and the plane's speed and distance.
  • Distance Calculation: The sound traveled for 60 seconds at 330 \( ext{m/s} \), giving a calculated distance of 19800 meters.
  • Plane's Path: The plane, moving at 412.5 \( ext{m/s} \), covers a distance of 24750 meters in the same time.
When computing altitude \( h \), where the altitude represents the vertical distance, we initially faced a mismatch, showing a recalibration necessity. Ensuring all sides align properly with scenario descriptions helps avoid pitfalls in calculations. An error in distance conceptual accuracy or arithmetic alignment is critical to note and resolve.

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