Question

A man fires a bullet standing b/w two cliffs. First echo is heard after 38 and the second echo is heard after 58. If the velocity of sound is $300\,{\text{m/s}}$ then the distance b/w the cliff is
A. $660\,{\text{m}}$
B. $14400\,{\text{m}}$
C. $1320\,{\text{m}}$
D. $1950\,{\text{m}}$

Answer 1

Use the formula for speed of an object. This formula gives the relation between speed of an object, distance travelled by the object and time required to travel the distance. Using this formula, calculate the distance between the man and first cliff and man and second cliff. Then calculate the total distance between the two cliffs by taking the addition of these two distances.

Formula used:
The speed $v$ of an object is given by
$v = \dfrac{d}{t}$ …… (1)
Here, $d$ is distance travelled by the object and $t$ is the time required to cover the same distance.

Complete step by step answer:
We have given that a man fires a bullet between the two cliffs. He heard the first echo 38 seconds after firing the bullet and then heard the second echo after 58 seconds after the firing of the bullet.
${t_1} = 38\,{\text{s}}$
${t_2} = 58\,{\text{s}}$
This shows that when the man fires the bullet the sound from firing of the bullet travels towards both the cliffs between which the man is standing and then reflects back to the man after reflection from the cliffs.
The velocity of the sound is $300\,{\text{m/s}}$.
$v = 300\,{\text{m/s}}$
We have asked to calculate the distance between the two cliffs.
Let ${d_1}$ be the distance of the first cliff from the man and ${d_2}$ be the distance of the second cliff from the man.The distance travelled by the sound when the echo is heard is twice the distance between the man and the cliff as the sound first hits the cliff and then returns back towards the man.Let us first calculate the distance between the man and first cliff.Rewrite equation (1) for the speed of the sound while travelling and coming back to the man for the first cliff.
$v = \dfrac{{2{d_1}}}{{{t_1}}}$
$\Rightarrow {d_1} = \dfrac{{v{t_1}}}{2}$
Substitute $300\,{\text{m/s}}$ for $v$ and $38\,{\text{s}}$ for ${t_1}$ in the above equation.
$\Rightarrow {d_1} = \dfrac{{\left( {300\,{\text{m/s}}} \right)\left( {38\,{\text{s}}} \right)}}{2}$
$\Rightarrow {d_1} = 5700\,{\text{m}}$
Hence, the distance between the man and the first cliff is $5700\,{\text{m}}$.
Let us now first calculate the distance between the man and second cliff.Rewrite equation (1) for the speed of the sound while travelling and coming back to the man for the second cliff.
$v = \dfrac{{2{d_2}}}{{{t_2}}}$
$\Rightarrow {d_2} = \dfrac{{v{t_2}}}{2}$
Substitute $300\,{\text{m/s}}$ for $v$ and $58\,{\text{s}}$ for ${t_2}$ in the above equation.
$\Rightarrow {d_2} = \dfrac{{\left( {300\,{\text{m/s}}} \right)\left( {58\,{\text{s}}} \right)}}{2}$
$\Rightarrow {d_2} = 8700\,{\text{m}}$
Hence, the distance between the man and the second cliff is $8700\,{\text{m}}$.

The distance $d$ between the two cliffs is the sum of the distance ${d_1}$ of man from the first cliff and the distance ${d_2}$ of man from the second cliff.
$d = {d_1} + {d_2}$
Substitute $5700\,{\text{m}}$ for ${d_1}$ and $8700\,{\text{m}}$ for ${d_2}$ in the above equation.
$d = \left( {5700\,{\text{m}}} \right) + \left( {8700\,{\text{m}}} \right)$
$\therefore d = 14400\,{\text{m}}$
Therefore, the distance between the two cliffs is $14400\,{\text{m}}$.

Hence, the correct option is B.

Note: The students should be careful while calculating the distance between the man and cliffs because the distance travelled by the sound when the first echo is heard is not equal to the distance between the man and cliff but it is equal to twice the distance between the man and cliff as the sound travels to the cliff and then comes back to man.