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(C) Let the total height be $H$. Since the stone is dropped,initial velocity $u = 0$. The time taken to reach the ground is $t = 5 \ s$. Using the equ

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(C) Let the total height be $H$. Since the stone is dropped,initial velocity $u = 0$. The time taken to reach the ground is $t = 5 \ s$. Using the equation $H = \frac{1}{2}gt^2$,we get:$H = \frac{1}{2} 	imes 10 	imes (5)^2 = 125 \ m$.In the first $3 \ s$,the distance covered is $h_1 = \frac{1}{2} 	imes 10 	imes (3)^2 = 45 \ m$.The remaining distance to be covered is $h_2 = H - h_1 = 125 - 45 = 80 \ m$.When the stone is released again after being stopped,it starts from rest $(u = 0)$ and covers the remaining distance $h_2$ in time $t'$.$h_2 = \frac{1}{2}gt'^2 \Rightarrow 80 = \frac{1}{2} 	imes 10 	imes t'^2$.$80 = 5t'^2 \Rightarrow t'^2 = 16 \Rightarrow t' = 4 \ s$.The total time taken is the initial $3 \ s$ plus the time $t'$ taken to cover the remaining distance.Total time $= 3 \ s + 4 \ s = 7 \ s$.

$A$ stone is dropped from a certain height and reaches the ground in $5 \ s$. It is stopped after $3 \ s$ of its fall and then released again. The total time taken by the stone to reach the ground will be .... $s$

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