A stone is dropped from the top of a tower 500 M high into a pond of water at the base of the tower. When is the splash heard at the top? Given \[g=10\text{ m/}{{\text{s}}^{2}}\] and speed of sound \[=340\text{ m/s}\]

Hint: In this question First we have to calculate the time to reach the stone at the pond and the time in which sound reaches at the top. Then we will add both times and it will give the required answer.

Complete step by step answer:
Let the time to hear the splash be t seconds.
Given, Height of tower H = 500 M.
\[g=10\text{ m/}{{\text{s}}^{2}}\] and speed of sound \[=340\text{ m/s}\]
When stone reaches towards the pond there will be the gravitational acceleration.
By motion's third Equation:
\[s=u{{t}_{1}}+\dfrac{1}{2}a{{t}_{1}}^{2}\]
Where \[S=H=500\text{ M}\]
\[u=0,a=g\] (along the gravitational acceleration)
Or towards earth
\[{{t}_{1}}=\] time taken to reach the ground
\[\Rightarrow 500=0+\dfrac{1}{2}\times 10\times {{t}_{1}}^{2}\]
\[\Rightarrow {{t}_{1}}=\sqrt{100}=10\text{ sec}\]
Time taken by sound to cover 500 M distance is \[{{t}_{2}}\]
So,
\[{{t}_{2}}=\dfrac{\text{Distance to travel}}{\text{speed of sound}}\]
\[\Rightarrow {{t}_{2}}=\dfrac{500}{340}=1.47\text{ sec}\]
Therefore,
Total time taken to hear the sound of splash
\[t={{t}_{1}}+{{t}_{2}}\]
\[\Rightarrow t=10+1.47=11.47\text{ sec}\]

Note:
When stone travels in the direction of earth’s center (surface) then it will move with positive gravitational acceleration.