Question 12 Marks
A stone is dropped into a well and its splash is heard at the mouth of the well after an interval of 1.45 s . Find the depth of the well. Given that velocity of sound in air at room temperature is equal to $332 ms^{-1}$.
Answer
View full question & answer→Let $h$ be the depth of the well. Then time $t_1$ taken by the stone to fall into well under gravity is given by $h=0+\frac{1}{2} g t_1^2$ or $t _1=\sqrt{\frac{2 h}{g}}$
Time taken for the splash to travel height h is given by $t_2=\frac{h}{v}$
where $v=$ velocity of sound
But $t _1+ t _2=1.45 s$
$\therefore \quad \sqrt{\frac{2 h}{g}}+\frac{h}{v}=1.45$
or $\sqrt{\frac{2 h}{9.8}}+\frac{h}{332}=1.45$
On solving, $h =9.9 m$.
Time taken for the splash to travel height h is given by $t_2=\frac{h}{v}$
where $v=$ velocity of sound
But $t _1+ t _2=1.45 s$
$\therefore \quad \sqrt{\frac{2 h}{g}}+\frac{h}{v}=1.45$
or $\sqrt{\frac{2 h}{9.8}}+\frac{h}{332}=1.45$
On solving, $h =9.9 m$.