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"^(-1)`
Time after which first echo is heard, `t_1 = 1.5s`
Time after which second echo is heard, `t_2 = 2.5 s`
Let A and B represent two parallel cliffs and O be the positions of the man stationed between these cliffs as shown in figure. 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 the 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,
`2d_1 = vt_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 `2d_2`.
or `d_2 = (vt_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"`