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A stone is dropped down to the water from a hill of $500 m$ in height. After what time will the splash be heard at the top of the hill? Take $g = 10 m / se c^{2}$ . The speed of sound in air is $340 m / \sec$ .

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10.53 \sec

11.47 \sec

15.33 \sec

18.12 \sec

answer is B.

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Detailed Solution

Total time taken to hear the sound of splash is given by

11.47 \sec

.
Height of the tower

S = 500 m

Initial speed of the stone

u = 0

Acceleration due to gravity,

g = 10 m / s^{2}

Initial velocity of the stone,

u = 0

(since the stone is initially at rest)
Time taken by the stone to fall to the base of the tower,

t_{1}

According to the second equation of motion:

S = u t_{1} + \frac{1}{2} g {t_{1}}^{2}

\Rightarrow 500 = 0 \times t_{1} + \frac{1}{2} \times 10 \times {t_{1}}^{2}

\Rightarrow

{t_{1}}^{2} = 100

\Rightarrow

t_{1} = 10 s

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

t_{2} = \frac{500}{340}

= 1.47 s

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

t = t_{1} + t_{2}

= 10 + 1.47

= 11.47 s .

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