# Q.13.  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,         $g=10\: m\: s^{-2}$ and speed of sound = $340\: m\: s^{-1}$.

D Divya Prakash Singh

Given the height of the tower, $s =500\ m$,

and the velocity of the sound, $v = 340\ m/s$,

Acceleration due to gravity, $g = 10\ ms^{-2}$

Let the initial velocity of the stone, $u = 0$, as the stone is initially at rest.

Let the time taken by the stone to fall to the base of the tower be $t_{1}$

Now, according to the IInd Equation of Motion:

$s = ut_{1}+\frac{1}{2}gt_{1}^2$

$500 = 0\times t_{1}+\frac{1}{2}\times10\times t_{1}^2$

$t_{1}^2 = 100$

$t_{1} = 10 sec$.

Now, the time taken by the sound to reach the top from the base of the tower,

$t_{2} = \frac{500}{340}sec =1.47sec$

Therefore, the splash is heard at the top after time, $t$

Calculating $t = t_{1}+t_{2} = 10+1.47 = 11.47sec.$

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