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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}.  

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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.


Posted by

Divya Prakash Singh

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