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\(\bullet$$\bullet\) You (height of your eyes above the water, 1.75 \(\mathrm{m}\) ) are standing 2.00 \(\mathrm{m}\) from the edge of a 2.50 -m-deep swimming pool. You notice that you can barely see your cell phone, which went missing a few minutes before, on the bottom of the pool. How far from the side of the pool is your cell phone?

Short Answer

Expert verified
1.18 meters from the side of the pool.

Step by step solution

01

Visualize the Problem

Imagine yourself standing on the edge of a swimming pool. Your eyes are at a height of 1.75 m above the ground, and the depth of the pool is 2.50 m. You are 2.00 m away from the edge of the pool.
02

Understand the Geometry

Visualize the situation as a right triangle, where your line of sight forms the hypotenuse. The vertical side of the triangle is the height of the eye level above the bottom of the pool (1.75 m + 2.50 m = 4.25 m), and the horizontal side is the distance to the phone from the edge of the pool.
03

Apply Geometry to Find the Distance

Using similar triangles, the ratios of corresponding sides are equal. Let the distance x from the edge of the pool to the phone be the horizontal side of the smaller triangle formed. Your height from the ground to eye level (1.75 m) is the height of the smaller triangle, and your distance from the pool (2.00 m) is the base of this smaller triangle.The equation is: \[ \frac{x}{2.00} = \frac{2.50}{4.25} \]
04

Solve the Equation

Simplify and solve the proportion: \[ x = \frac{2.50}{4.25} \times 2.00 \]Calculate this product to find the value of x.
05

Calculate the Distance

Perform the arithmetic calculation: \( \frac{2.50}{4.25} \approx 0.5882 \)Therefore, \( x = 0.5882 \times 2.00 = 1.1764 \) m.
06

Conclusion

The cell phone is located approximately 1.18 meters from the side of the pool.

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Key Concepts

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

Right Triangle
In geometry, a right triangle is a triangle in which one of the angles is exactly 90 degrees. This type of triangle is fundamental in various applications, especially in physics and engineering.
The special property of a right triangle is that the square of the length of the hypotenuse (the longest side, opposite the right angle) equals the sum of the squares of the lengths of the other two sides. This relationship is famously known as the Pythagorean theorem and can be expressed as: \[ c^2 = a^2 + b^2 \] where \( c \) is the hypotenuse, and \( a \) and \( b \) are the other two sides. This theorem helps in calculations and solving real-world problems like locating objects.
In the swimming pool problem, the situation is represented by a right triangle where your line of sight from the edge of the pool to the cell phone is the hypotenuse. This geometric visualization sets the foundation for using geometric concepts to find distances.
Similar Triangles
Similar triangles are triangles that have the same shape but may differ in size. Each corresponding angle in similar triangles is equal, and the ratios of the corresponding sides are constant. This property is essential for solving many geometry problems, as it allows for the use of proportional reasoning to calculate unknown lengths.
For instance, in the swimming pool problem, two similar triangles are used to find the cell phone's location. One triangle is the larger overall triangle created by your height from the bottom of the pool to your eye level, and the distance from you to the pool's edge. The smaller, similar triangle is below your line of sight and shares a base parallel to the hypothetical line extending from your eyes to the pool's bottom in relation to the cell phone.
The proportions allow us to set up the equation \( \frac{x}{2.00} = \frac{2.50}{4.25} \), where \( x \) is the distance from the pool's side to the phone. This approach demonstrates how similar triangles simplify and clarify the problem, making distance calculation more accessible.
Swimming Pool Problem
The swimming pool problem is an intriguing application of physics and geometry combined. It revolves around practical visualization and calculation using geometric concepts to find missing objects.
This scenario places you at the edge of a pool, looking for a cell phone that has accidentally sunk to the bottom. Your height and distance from the pool create a geometric setup where understanding the relationships between lines of sight and distance aids in finding the phone. By visualizing the situation as a series of triangles, these mathematical tools allow identification of the phone's position under the water.
The swimming pool problem embodies the beauty of geometry in real life; it shows how mathematical principles help solve everyday issues. This kind of problem trains intuition and problem-solving skills, crucial in both academic studies and practical situations.
Distance Calculation
Calculating distances in geometry often involves practical techniques using known geometric properties. In this context, we're applying proportions derived from similar triangles to determine the unknown distance between the pool’s edge and the phone.
The actual calculation begins by establishing the relationship with the equation \( \frac{x}{2.00} = \frac{2.50}{4.25} \). Solving for \( x \) involves basic arithmetic where:
  • First, calculate the ratio: \( \frac{2.50}{4.25} \approx 0.5882 \).
  • Then multiply by the known base of the triangle: \( 0.5882 \times 2.00 \ = 1.1764 \) m.
This calculated distance gives a practical gauge for locating the cellphone below the water. Distance calculation, as presented here, emphasizes the power of combining simple geometric technique and relative measurements to directly solve an otherwise invisible problem.

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Most popular questions from this chapter

\(\bullet$$\bullet\) A small helium-neon laser emits red visible light with a power of 3.20 \(\mathrm{mW}\) in a beam that has a diameter of 2.50 \(\mathrm{mm}\) . (a) What are the amplitudes of the electric and magnetic fields of the light? (b) What are the average energy densities associated with the electric field and with the magnetic field? (c) What is the total energy contained in a 1.00 \(\mathrm{m}\) length of the beam?

\(\bullet\) Unpolarized light with intensity \(I_{0}\) is incident on an ideal polarizing filter. The emerging light strikes a second ideal polarizing filter whose axis is at \(41.0^{\circ}\) to that of the first. Determine (a) the intensity of the beam after it has passed through the second polarizer and (b) its state of polarization.

\(\bullet$$\bullet\) A beaker with a mirrored bottom is filled with a liquid whose index of refraction is \(1.63 .\) A light beam strikes the top surface of the liquid at an angle of \(42.5^{\circ}\) from the normal. At what angle from the normal will the beam exit from the liquid after traveling down through it, reflecting from the mirrored bottom, and returning to the surface?

\(\bullet\) Ultraviolet radiation. There are two categories of ultraviolet light. Ultraviolet A (UVA) has a wavelength ranging from 320 \(\mathrm{nm}\) to 400 nm. It is not so harmful to the skin and is necessary for the production of vitamin D. UVB, with a wavelength between 280 \(\mathrm{nm}\) and \(320 \mathrm{nm},\) is much more dangerous, because it causes skin cancer. (a) Find the frequency ranges of UVA and UVB. (b) What are the ranges of the wave numbers for UVA and UVB?

\(\bullet\) The intensity at a certain distance from a bright light source is 6.00 \(\mathrm{W} / \mathrm{m}^{2} .\) Find the radiation pressure (in pascals and in atmospheres) on (a) a totally absorbing surface and (b) a totally reflecting surface.

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