A person standing between two vertical cliffs and `680 m` away from the nearst cliff, shouted. He heard the first echo after `4 s` and the second echo `3 s` later. Calculate (a) the speed of sound in air and (b) distance between the two cliffs.

Answer

(a) 340 m//s
(b) 1870 m.

Explanation

Here, `t_1 = 4 s and t_2 = 3 s, d_1 = 680 m`
(a) As `d_1 = (vt)/(2), v = (2 d_1)/(t_1) = (2 xx 680)/(4) = 340 m//s`
(b) As `d_2 = (v(t_1 + t_2))/(2) = (340(4 + 3))/(2) = 1190 m`
Distance between two cliffs, `d = d_1 + d_2 = 680 + 1190 = 1870 m`.