Question

2) A person stands in between two parallel cliffs which are 99 m apart. He fires a gun and hears two successive echoes after 0.2 s and 0.4 s. Calculate : (a) the distance of the person from the nearer cliff (b) speed of sound. $\left[\left(a\right)33m\left(b\right)330m{s}^{-1}\right]$

Answer 1

Dear student,
2) velocity of sound, v=2d/t

v = 2 x 99m/0.6s =330 m/s

For nearer cliff= v = 2d/t
v = 2x/0.2
330 = x/0.1
x= 33 m
Regards