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Question

(A) Let $t_{1}$ be the time taken by the stone to strike the surface of the water in the pond.Using the equation of motion $h = ut + \frac{1}{2}gt^{2}

Vedclass · Accepted Answer

$\sqrt{\frac{2h}{g}} + \frac{h}{v}$

Vedclass · Answer

(A) Let $t_{1}$ be the time taken by the stone to strike the surface of the water in the pond.Using the equation of motion $h = ut + \frac{1}{2}gt^{2}$:Since the stone is dropped,initial velocity $u = 0$.Therefore,$h = \frac{1}{2}gt_{1}^{2}$.Solving for $t_{1}$,we get $t_{1} = \sqrt{\frac{2h}{g}}$.Now,the time taken by the sound of the splash to travel from the surface of the water to the top of the tower is $t_{2} = \frac{h}{v}$,where $v$ is the speed of sound.The total time $t$ after which the splash is heard is the sum of the time taken by the stone to fall and the time taken by the sound to travel back up:$t = t_{1} + t_{2} = \sqrt{\frac{2h}{g}} + \frac{h}{v}$.

$A$ stone is dropped into a pond from the top of a tower of height $h$. If $v$ is the speed of sound in air,then the sound of the splash will be heard at the top of the tower after a time:

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