Question

A man standing in front of a vertical cliff is firing a gun. He hears the echo after $3s$. On moving closer to the cliff by a distance of $82.5m$, he fires again and hears the echo after $2.5s$. Calculate the distance of the cliff from the initial position of the man.

Answer 1

The speed of the sound is calculated by taking the ratio of the distance to the time taken. When the man moves, the distance is being changed, the time also varies. The speed will remain a constant. So we can equate this. Hence we can find the distance of the initial position of the man from the cliff. These all may help you to solve this question.

Complete step-by-step answer:
Let us assume that the distance of the cliff from the initial position of the man be given as $d$. It is travelling the to and fro distance. Therefore the total distance travelled be $2d$.

The time taken to travel this distance is given as,
$t=3s$
Therefore, the speed of travel can be found by the equation,
$S=\dfrac{d}{t}$
Substituting the values in it,
$S=\dfrac{2d}{3}$
When the man moves towards the cliff by a distance,
${d}'=82.5m$
Therefore the resultant distance between the cliff and the man is given as,
$d-{d}'=d-82.5$
The total distance will be twice of this. That is,
${{d}_{f}}=2\times \left( d-82.5 \right)$
The time taken will be,
${{t}_{f}}=2.5s$
Therefore the speed will be given as,
$S=\dfrac{2\left( d-82.5 \right)}{2.5}$
These two speeds will be equivalent. That is,
$S=\dfrac{2\left( d-82.5 \right)}{2.5}=\dfrac{2d}{3}$
Simplifying this equation,
$5d=6d-495$
Therefore the distance between the cliff and the initial position of man will be,
$d=495m$
Therefore the answer has been obtained.

Note: Speed of the body or a wave is a scalar quantity. The direction of the quantity is not being considered here. The speed is given by the unit as meter per second. The velocity is a vector quantity different from the speed of the object.