/*! 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} Q7P A stone is dropped from the top ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A stone is dropped from the top of a cliff. The splash it makes when striking the water below is heard 2.7 s later. How high is the cliff?

Short Answer

Expert verified

The height of the cliff is \(33.24\;{\rm{m}}\).

Step by step solution

01

Understanding the concept of distance travelled by system during free fall

During the free-fall motion, only a gravitational force is generating acceleration over the system. The distance travelled by the system during free fall is a function of the initial speed of the object, time of travel, and gravitational acceleration.

02

Given data

The time of striking heard is \(T = 2.7\;{\rm{s}}\).

03

Evaluating the value of travel time of sound to reach the top of cliff and time taken by stone to reach the water surface

The standard value for the speed of sound in air from Table-12.1 is \(v = 343\;{\rm{m/s}}\), and the standard value for gravitational acceleration is \(g = 9.81\;{\rm{m/}}{{\rm{s}}^2}\).

The travel time of sound to reach the top of cliff is calculated below:

\({t_2} = \frac{h}{v}\)

Here, h is the height of cliff.

The time taken by stone to reach the water surface is calculated below:

\(\begin{aligned}{l}T = {t_1} + {t_2}\\{t_1} = T - {t_2}\\{t_1} = T - \left( {\frac{h}{v}} \right)\end{aligned}\)

04

Evaluating the height of a cliff by using the Newton’s equation of motion

The height of cliff is calculated below:

\(\begin{aligned}{c}h = u{t_1} + \frac{1}{2}g{\left( {{t_1}} \right)^2}\\h = \left( 0 \right){t_1} + \frac{1}{2}g{\left[ {T - \left( {\frac{h}{v}} \right)} \right]^2}\\{h^2} - 2v\left( {\frac{v}{g} + T} \right)h + {T^2}{v^2} = 0\end{aligned}\)

Here, u is the initial speed of stone having a value of zero, g is the gravitational acceleration having a standard value of \(9.81\;{\rm{m/}}{{\rm{s}}^2}\).

Substitute the values in the above equation.

\(\begin{aligned}{c}{h^2} - 2\left( {343\;{\rm{m/s}}} \right)\left( {\frac{{343\;{\rm{m/s}}}}{{9.81\;{\rm{m/}}{{\rm{s}}^2}}} + 2.7\;{\rm{s}}} \right)h + {\left( {2.7\;{\rm{s}}} \right)^2}{\left( {343\;{\rm{m/s}}} \right)^2} = 0\\{h^2} - 25837.72h + 857661.21 = 0\end{aligned}\)

On solving the above quadratic equation,

\(h = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\)

Here, a, b and c are the equation constants for \(a{h^2} + bh + c = 0\).

Substitute the values in the above equation.

\(\begin{aligned}{c}h = \frac{{ - \left( { - 25837.72} \right) \pm \sqrt {{{\left( { - 25837.72} \right)}^2} - 4\left( 1 \right)\left( {857661.21} \right)} }}{{2\left( 1 \right)}}\\h = 33.24\;{\rm{m}}\end{aligned}\)

The largest root is impossible to consider because it will consume more time than 2.7 seconds to fall at that distance.

Hence, the height of cliff is \(33.24\;{\rm{m}}\).

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 91Ó°ÊÓ!

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

Do you expect an echo to return to you more quickly on a hot day or a cold day?

(a) Hot day.

(b) Cold day.

(c) Same on both days.

Question: A tuning fork is set into vibration above a vertical open tube filled with water (Fig 12-40). The water level is allowed to drop slowly. As it does so, the air in the tube above the water level is heard to resonate with the tuning fork when the distance from the tube opening to the water level is 0.125 m and again at 0.395 m. What is the frequency of the tuning fork?

FIGURE 12–40 Problem 79.

Question: (III) When a player’s finger presses a guitar string down onto a fret, the length of the vibrating portion of the string is shortened, thereby increasing the string’s fundamental frequency (see Fig. 12–36). The string’s tension and mass per unit length remain unchanged. If the unfingered length of the string is l= 75.0 cm, determine the positions x of the first six frets, if each fret raises the pitch of the fundamental by one musical note compared to the neighboring fret. On the equally tempered chromatic scale, the ratio of frequencies of neighboring notes is 21/12.

Figure 12-36

In which of the following is the wavelength of the lowest vibration mode the same as the length of the string or tube?

(a) A string.

(b) An open tube.

(c) A tube closed at one end.

(d) All of the above.

(e) None of the above.

Question: (II) A tight guitar string has a frequency of 540 Hz as its third harmonic. What will be its fundamental frequency if it is fingered at a length of only 70% of its original length?

See all solutions

Recommended explanations on Physics Textbooks

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.