A man fires a shot and hears an echo from a cliff after `2 s`. He walks `85 m` towards the cliff and the echo of a second shot is now heard after `1.4 s`. What is the velocity of sound and how far was the man from the cliff when he first heard the echo ?

Explanation

Let the man stationed at A at a distance d from the cliff. If the cliff is denoted by C, then AC= d as shown in figure.

The echo of the first shot will be heard if the sound goes from A to C and comes back to A. In the process, it travels a distance 2d.
Since a time of 2 s is taken by the first echo to be heard,
`2d = vt = v xx 2 = 2v" "(v " is speed of sound")`
or `d = v" "...(i)`
After walking 85 m, the man comes to B where BC `= ( d- 85)`m. For the echo of the second shot to be heard, sound has to go from B to C and back to the man B, i.e., it has to travel a distance `2BC = 2 (d- 85) m = (2d- 170)m.`
Since echo of the second shot is heard after a time of `1.5`s
`2d - 170 = v xx 1.5 = (3v)/(2)" "...(ii)`
From equation (i) and (ii),
or `d = 340` m
Speed of sound, `v = d = 340"m s"^(-1)`