Question

A stone is dropped into a well the splash of water is heard after 3.91s. .If the well is 67.6m find the velocity of sound

Answer 1

Answer:

S = 1/2 g t1^2  = 67.6    time for stone to hit water

t1 = (2 S / g)^1/2 = (2 * 67.6 / 9.8)^1/2 = 3.71 s

t2 = T - t1 = 3.91 - 3.71 = .2 s      time for sound to reach observer

V = S / t2 = 67.6 / .2 = 338 m/s     speed of sound

Answer 2

The velocity of sound is 338 m/s.

Explanation:

A stone is dropped into a well the splash of water is heard after 3.91 s. If the well is 67.6 m. We need to find the velocity of sound.

Let t is the time for stone to hit water. Using second equation of motion as :

$s=ut+\dfrac{1}{2}gt^2\\\\u=0\\\\s=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2s}{g}} \\\\t=\sqrt{\dfrac{2\times 67.6}{9.8}} \\\\t=3.71\ s$

Let t' is the time for sound to reach observer. So,

t' = T -t

t' = 3.91 s - 3.71 s = 0.2 s

The speed of sound is given by :

$v=\dfrac{s}{t}\\\\v=\dfrac{67.6\ m}{0.2\ s}\\\\v=338\ m/s$

So, the velocity of sound is 338 m/s.

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