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(A) The total time taken to hear the splash is the sum of the time taken by the stone to fall $(t_1)$ and the time taken by the sound to travel back t

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$8.7$

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(A) The total time taken to hear the splash is the sum of the time taken by the stone to fall $(t_1)$ and the time taken by the sound to travel back to the top $(t_2)$.$1$. Time taken by the stone to reach the water $(t_1)$:Using the equation of motion $s = ut + \frac{1}{2}gt^2$,where $u = 0$,$s = 300 \; m$,and $g = 9.8 \; m s^{-2}$:$300 = 0 + \frac{1}{2} 	imes 9.8 	imes t_1^2$$t_1^2 = \frac{600}{9.8} \approx 61.22$$t_1 = \sqrt{61.22} \approx 7.82 \; s$$2$. Time taken by the sound to reach the top $(t_2)$:Using $t_2 = \frac{	ext{distance}}{	ext{speed}}$:$t_2 = \frac{300}{340} \approx 0.88 \; s$$3$. Total time $(t)$:$t = t_1 + t_2 = 7.82 + 0.88 = 8.7 \; s$.

$A$ stone dropped from the top of a tower of height $300 \; m$ splashes into the water of a pond near the base of the tower. When is the splash heard at the top,given that the speed of sound in air is $340 \; m s^{-1}$ (in $; s$)? $(g = 9.8 \; m s^{-2})$

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