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(B) Let the total time of motion be $n$ seconds. The distance travelled in the $n^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.For a freel

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$2$

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(B) Let the total time of motion be $n$ seconds. The distance travelled in the $n^{th}$ second is given by $S_n = u + \frac{a}{2}(2n - 1)$.For a freely falling body,initial velocity $u = 0$ and acceleration $a = g = 10 \ m \ s^{-2}$.The distance travelled in the last second ($n^{th}$ second) is $S_n = 0 + \frac{10}{2}(2n - 1) = 5(2n - 1)$.The distance travelled in the last but one second ($(n-1)^{th}$ second) is $S_{n-1} = 0 + \frac{10}{2}(2(n-1) - 1) = 5(2n - 3)$.Given that the distance in the last but one second is $5 \ m$,we have $5(2n - 3) = 5$.$2n - 3 = 1 \implies 2n = 4 \implies n = 2 \ s$.

If the distance travelled by a freely falling body in the last but one second of its motion is $5 \ m$,then the total time of motion is (Acceleration due to gravity $g = 10 \ m \ s^{-2}$) (in $s$)

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