Question

A man standing in front of a vertical cliff fire a gun. He hears the echo after $3\mathrm{s}$. On moving closer to the cliff by $82.5\mathrm{m}$, he fires again. This time, he hears the echo after $2.5\mathrm{s}$. Calculate the distance of the cliff from the initial position of the man.

Answer 1

Step 1: Given data:

The initial distance of the person from the cliff is $D$

The final distance of the person from the cliff is $D-82.5m$

Time after which the echo is heard from his initial position, $T_1=3s$

Time after which the echo is heard from his initial position, $T_2=2.5s$

Step 2: Calculating the distance of the cliff from the final position of the man:
Let $v$ be the speed of sound

Speed of sound, $v==\frac{2\times dis\tan ce}{T_2}=\frac{2(D-82.5)}{2.5}....\left(1\right)$

Step 3: Calculating the distance of the cliff from the initial position of the man

We know that the total distance traveled by the sound will be twice the distance between the man and the cliff.

Thus, the total distance that the sound travels before reaching the person is $2D$
Speed of sound, $v=\frac{2\times dis\tan ce}{T_1}=\frac{2D}{3}....\left(2\right)$
Since the speed of sound is the same before and after the movement of the man.

Thus, from equation $\left(1\right)$ and equation $\left(2\right)$ we have,

$\frac{2D}{3}=\frac{2(D-82.5)}{2.5}$

$$\begin{gathered} 5D=6D-495 \\ D=495m \end{gathered}$$

Hence, the distance of the cliff from the initial position of the man is $495m$.