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(B) The total time $t$ 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

Vedclass · Accepted Answer

$11.47$

Vedclass · Answer

(B) The total time $t$ 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 for the stone to fall $(t_1)$:Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$,$a = g = 10\, m\, s^{-2}$,and $s = 500\, m$:$500 = 0 + \frac{1}{2} 	imes 10 	imes t_1^2$$500 = 5 	imes t_1^2$$t_1^2 = 100$$t_1 = 10\, s$.$2$. Time taken for the sound to travel back $(t_2)$:Using the formula $t_2 = \frac{	ext{distance}}{	ext{speed}}$:$t_2 = \frac{500}{340} \approx 1.47\, s$.$3$. Total time $(t)$:$t = t_1 + t_2 = 10 + 1.47 = 11.47\, s$.

$A$ stone is dropped from the top of a tower $500\, m$ high into a pond of water at the base of the tower. When is the splash heard (in $sec$) at the top?
Given,$g = 10\, m\, s^{-2}$ and speed of sound $= 340\, m\, s^{-1}$.

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$A$ stone is dropped from the top of a tower $500\, m$ high into a pond of water at the base of the tower. When is the splash heard (in $sec$) at the top?Given,$g = 10\, m\, s^{-2}$ and speed of sound $= 340\, m\, s^{-1}$.