A man stationed between two parallel cliffs fires a gun. He hears the first echo after `1.5 s` and the next after `2.5 s`. What is the distance between the cliffs and when does the hear the third echo ? Take the speed of sound in air as `340 m//s`.

Explanation

Here, velocity of sound in air, `v = 340 m//s`
time after which first echo is heard, `t_1 = 1.5 s`
time after which second echo is heard, `t_2 = 2.5 s`
Let `A and B` represent two parallel cliffs and `O` be the position of the man stationed between these cliffs as shown in (Fig. 6.23). Let `d_1` and `d_2` be the distances of the man from these cliffs A and B respectively.
Reflection from cliff `A`
Time `(t_1)` after which first echo is heard = time taken by sound to travel from O to A and back to O again, i.e., a distance `2 d_1`.
As distance travelled by sound = speed of sound `xx` time,
`2 d_1 = v t_1` or `d_1 = (vt_1)/(2) = (340 xx 1.5)/(2) = 255 m`.
Reflection from cliff `B`
Time `(t_2)` after which the second echo is heard = time taken by sound to travel from O to B and back to O again, i.e., a distance `2 d_2`.
Clearly, `2 d_2 = v t_2` or `d_2 = (v t_2)/(2) = (340 xx 2.5)/(2) = 425 m`
Distance between the two cliffs, `d = d_1 + d_2 = 255 m + 425 m = 680 m`
The third echo is heard when sound after reflection from cliff `A` goes to cliff `B` and then after reflection from it reaches `O`.
Time taken for the third echo to be heard, `t = t_1 + t_2 = 1.5 s + 2.5 s = 4 s`.
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