Question

A stone dropped from the top of a tower of height $125m$ splashes into a pond of water at its base. When will the sound of the splash be heard at the top?
(Take velocity of sound = $340m{s}^{-1}$ and g = $10m{s}^{-2}$)

• A. 6.62 s
• B. 5.368 s
• C. 5.95 s
• D. 6.79 s

Answer 1

The correct option is B

5.368 s

Given, distance, s = 125 m, velocity of sound, v = $340m{s}^{-1}$ , g = $10m{s}^{-2}$

Time after which the splash is heard at the top is equal to the sum of the time $t_1$ taken by the stone to fall down and the time $t_2$ taken by the sound to travel from bottom to top.
Using, $s=ut+\frac{1}{2}a{t}^2$, where s = displacement, u = initial velocity, t = time, a = accelaration (here g)
We get, $s=\frac{1}{2}g{t}_1^2$(since initial velocity,u = 0)
$t_1=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2\times 125}{10}}\Rightarrow t_1$ = $5s$
Again $t_2=\frac{heightoftower}{velocityofsound}$
$t_2=\frac{125}{340}=0.368s$
Total time $=t_1+t_2=(5+0.368)s=5.368s$