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(C) Let the total height be $H$. The stone is dropped from rest,so $u = 0$.Using the equation of motion $H = \frac{1}{2}gt^2$,for $t = 5\, s$,we get $

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$4$

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(C) Let the total height be $H$. The stone is dropped from rest,so $u = 0$.Using the equation of motion $H = \frac{1}{2}gt^2$,for $t = 5\, s$,we get $H = \frac{1}{2}g(5)^2 = \frac{25}{2}g$.Distance covered in the first $3\, s$ is $h_1 = \frac{1}{2}g(3)^2 = \frac{9}{2}g$.Remaining distance $h_2 = H - h_1 = \frac{25}{2}g - \frac{9}{2}g = \frac{16}{2}g = 8g$.When the stone is stopped after $3\, s$,its velocity at that instant is $v = gt = 3g$. However,it is then allowed to fall again from rest (as it was stopped),so for the remaining distance $h_2$,the initial velocity $u' = 0$.Using $h_2 = \frac{1}{2}gt'^2$,where $t'$ is the time to cover the remaining distance:$8g = \frac{1}{2}gt'^2$$16 = t'^2$$t' = 4\, s$.

$A$ stone is dropped from a certain height and reaches the ground in $5\, s$. If the stone is stopped after $3\, s$ of its fall and then allowed to fall again,then the time taken by the stone to reach the ground for the remaining distance is........$s$.

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