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Chapter 2: Problem 84

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help." When she has fallen for 3.0 \(\mathrm{s}\) , she hears the echo of her shout from the valley floor below. The speed of sound is 340 \(\mathrm{m} / \mathrm{s}\) . (a) How tall is the cliff? \((b)\) If air resistance is neglected, bow fast will she be moving just before she hits the ground? (Her actual speed will be less than this, due to air resistance.)

Short Answer

Expert verified
The cliff is about 44.1 m tall, and her speed is 29.4 m/s.

Step by step solution

01

Understand the Problem

A teacher falls from a cliff and after 3 seconds hears the echo of her shout. We need to find the height of the cliff and her speed before hitting the ground. The sound speed is 340 m/s.
02

Calculate Time for Sound to Travel

When she falls, it takes some time for her shout to reach the ground and bounce back, totaling the 3 seconds. Let's say the time taken to fall is \(t_1\) and the time taken for the sound to return is \(t_2\). So, \(t_1 + t_2 = 3\).
03

Express \(t_2\) in Terms of \(h\)

The sound travels the tall distance \(h\) down and back up at 340 m/s. Thus, \(t_2 = \frac{2h}{340}\).
04

Find \(t_1\) from Fall Equation

Using the equation of motion \(h = \frac{1}{2}gt_1^2\) and substituting \(g = 9.8 \text{ m/s}^2\), we solve for \(t_1\). Rearrange to get \(t_1^2 = \frac{2h}{9.8}\).
05

Establish Two Equations

We equate and solve: 1) \(t_1^2 = \frac{2h}{9.8}\) and 2) \(t_1 + \frac{2h}{340} = 3\).
06

Solve for \(h\) and \(t_1\)

Substitute \(t_1\) from the first equation into the second, resulting in a quadratic equation in \(h\): \(\frac{2h}{9.8} + \frac{2h}{340} = 9\), solving to find \(h \approx 44.1 \text{ m}\).
07

Calculate Final Velocity

Use \(v = gt_1\) to find the final velocity of the teacher just before hitting the ground, substituting \(t_1 \approx \sqrt{\frac{2h}{9.8}} = \sqrt{\frac{2 \times 44.1}{9.8}} \approx 3 \text{ s}\). Thus, \(v = 9.8 \times 3 = 29.4 \text{ m/s}\).

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Key Concepts

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

Free Fall Motion
Free fall motion is when an object is moving only under the influence of gravity. This type of motion eliminates other forces such as air resistance. When an object is in free fall, its initial velocity is often zero, just as in the case of the teacher falling off the cliff.
Gravity accelerates the object at a constant rate, usually approximated as 9.8 m/s虏 on Earth. Hence, the object's speed increases by 9.8 m/s each second it falls.
This constant acceleration leads to specific equations, such as the one used above: \( h = \frac{1}{2}gt_1^2 \). This formula reveals that the distance \( h \) the teacher falls depends on the square of the time \( t \) she is in free fall, multiplied by half the gravitational acceleration.
Sound Echo Calculation
Sound echo calculation is essential for determining distances using sound waves. An echo occurs when sound waves bounce off a distant surface and return to the point of origin.
In the problem, an echo is heard when the teacher shouts "Help" and the sound travels down to the valley floor and back up to her. The total time for this round-trip is the time elapsed minus her falling time.
This is given as \( t_2 = \frac{2h}{340} \), where 340 m/s is the speed of sound. Here, \( h \) is the height of the cliff. These calculations can be complex but are important in various real-world applications like sonar and radar.
Equations of Motion
Equations of motion are used to describe the fundamental principles of an object's movement, containing variables like displacement, time, velocity, and acceleration.
In this exercise, equations are used to link time and height: \( t_1 + \frac{2h}{340} = 3 \) and \( t_1^2 = \frac{2h}{9.8} \).
These equations help solve problems by eliminating unknown values step-by-step. By plugging \( t_1 \) from one equation into another, a quadratic equation is formed for the height \( h \).
These mathematical representations are powerful tools for understanding real-life situations governed by the laws of physics.

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Most popular questions from this chapter

From Pillar to Post. Starting from a pillar, you run 200 \(\mathrm{m}\) east (the \(+x\) -direction) at an average speed of 5.0 \(\mathrm{m} / \mathrm{s}\) , and then run 280 \(\mathrm{m}\) west at an average speed of 4.0 \(\mathrm{m} / \mathrm{s}\) to a post. Calculate (a) your average speed from pillar to post and (b) your average velocity from pillar to post.

Touchdown on the Moon. A lunar lander is making its descent to Moon Base I (Fig. 2.40). The lander descends slowly under the retro-thrust of its descent engine. The engine is cut off when the lander is 5.0 \(\mathrm{m}\) above the surface and has a downward speed of 0.8 \(\mathrm{m} / \mathrm{s}\) . With the engine off, the lander is in free fall. What is the speed of the lander just before it touches the surface? The acceleration due to gravity on the moon is 1.6 \(\mathrm{m} / \mathrm{s}^{2}\) .

An alert hiker sees a boulder fall from the top of a distant cliff and notes that it takes 1.30 s for the boulder to fall the last third of the way to the ground. You may ignore air resistance. (a) What is the height of the cliff in meters? (b) If in part (a) you get two solutions of a quadratic equation and you use one for your answer, what does the other solution represent?

You are standing at rest at a bus stop. A bus moving at a constant speed of 5.00 m/s passes you. When the rear of the bus is 12.0 m past you, you realize that it is your bus, so you start to run toward it with a constant acceleration of 0.960 m/s. How far would you have to run before you catch up with the rear of the bus, and how fast must you be running then? Would an average college student be physically able to accomplish this?

Juggling Act. A juggler performs in a room whose ceiling is 3.0 \(\mathrm{m}\) above the level of his hands. He throws a ball upward so that it just reaches the ceiling. (a) What is the initial velocity of the ball? (b) What is the time required for the ball to reach the ceiling? At the instant when the first ball is at the ceiling, the juggler throws a second ball upward with two- thirds the initial velocity of the first. (c) How long after the second ball is thrown did the two balls pass each other? (d) At what distance above the juggler's hand do they pass each other?
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