A stone is dropped into a well that is 19.6,m | vedclass

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(B) The total time $t$ taken to hear the echo is the sum of the time taken by the stone to fall to the water surface $(t_1)$ and the time taken by the

Vedclass · Accepted Answer

$326.7$

Vedclass · Answer

(B) The total time $t$ taken to hear the echo is the sum of the time taken by the stone to fall to the water surface $(t_1)$ and the time taken by the sound to travel back to the top $(t_2)$.$t = t_1 + t_2$For the stone falling from rest,$h = \frac{1}{2}gt_1^2$,so $t_1 = \sqrt{\frac{2h}{g}}$.Given $h = 19.6\,m$ and $g = 9.8\,m/s^2$,$t_1 = \sqrt{\frac{2 	imes 19.6}{9.8}} = \sqrt{4} = 2.0\,s$.The time taken by sound to travel back is $t_2 = t - t_1 = 2.06\,s - 2.0\,s = 0.06\,s$.The velocity of sound $v$ is given by $v = \frac{h}{t_2} = \frac{19.6}{0.06} \approx 326.7\,m/s$.

$A$ stone is dropped into a well that is $19.6\,m$ deep. If the echo is heard after $2.06\,s$ of dropping the stone,calculate the velocity of sound in $m/s$.

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