A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower.
Question.
A stone is dropped from the top of a tower $500 \mathrm{~m}$ high into a pond of water at the base of the tower. When is the splash heard at the top ? Given, $g=10 \mathrm{~ms}^{-2}$; speed of sound $=$ $340 \mathrm{~m} / \mathrm{s} .$

Solution:

For the downward journey of stone,

Initial velocity, $\mathrm{u}=0$

Distance travelled, s = height of tower

= 500 m

Time of fall, $\mathrm{t}_{1}=?$

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

$\mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^{2}$

$s=u t+\frac{1}{2} g t^{2}$

or $500=0 \times \mathrm{t}_{1}+\frac{1}{2} \times 10 \times \mathrm{t}_{1}^{2}$

or $500=5 \mathrm{t}_{1}{ }^{2} \quad$ or $\quad \mathrm{t}_{1}{ }^{2}=100$

$\therefore \mathrm{t}_{1}=10 \mathrm{~s}$

For the sound travelling upward,

Time taken, $\mathrm{t}_{2}=?$

There is no effect of gravity on the propagation of sound i.e, always the formula of uniform motion is used for sound or any other wave.

Speed of sound, $v=\frac{s}{t_{2}}$

or $t_{2}=\frac{s}{v}=\frac{500}{340}=1.47 \mathrm{~s}$

Total time $=\mathrm{t}_{1}+\mathrm{t}_{2}=10+1.47=11.47 \mathrm{~s}$
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