Q. A man is standing between two parallel cliffs and fires a gun. If he hears first and second echoes after 1.5 s and 3.5 s respectively the distance between the cliffs is : (Velocity of sound in air $ =340m{{s}^{-1}} $ )
EAMCETEAMCET 2000
Solution:
Let the person is at point P. Time taken in first echo, $ {{t}_{1}}=1.5\,s $
$ \therefore $ $ x=\frac{V\times {{t}_{1}}}{2} $ $ =\frac{340\times 7.5}{2}=255\,m $ Similarly, $ y=\frac{V\times {{t}_{2}}}{2} $ $ =\frac{340\times 3.5}{2} $ $ =595\,m $ Distance between two cliffs $ =\text{ }x\text{ }+\text{ }y $ = 255 + 595 = 850 m
