A man standing between two parallel cliffs fires a gun. He hear two echos one after 3 second and the other after 5 second. The separation between the two cliffs is 1360 m, the speed of sound is

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Let the distance of the man from the first cliff be $d_{1}$ and from the second cliff be $d_{2}$ .

Total distance between the cliffs is $d_{1} + d_{2} = 1360$ m.

Time taken for the first echo, $t_{1} = 3$ s.

Time taken for the second echo, $t_{2} = 5$ s.

Let the speed of sound be $v$ .

Distance traveled by sound for the first echo is $2 d_{1} = v t_{1} = 3 v$ .

Distance traveled by sound for the second echo is $2 d_{2} = v t_{2} = 5 v$ .

Adding the two equations:

$2 d_{1} + 2 d_{2} = 3 v + 5 v$

$2 (d_{1} + d_{2}) = 8 v$

Substituting $d_{1} + d_{2} = 1360$ m:

$2 \times 1360 = 8 v$

$2720 = 8 v$

$v = \frac{2720}{8} = 340$ m/s.

Answer: $340$ m/s

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