A body falls from a height h = 200,m (at New | vedclass

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(C) For a body falling from rest under gravity,the distance $S$ travelled in time $t$ is given by $S = \frac{1}{2}gt^2$.The distance travelled in succ

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$1:3:5$

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(C) For a body falling from rest under gravity,the distance $S$ travelled in time $t$ is given by $S = \frac{1}{2}gt^2$.The distance travelled in successive equal time intervals $\Delta t$ starting from $t=0$ follows the ratio of odd numbers.Let the time interval be $\Delta t = 2\,s$.Distance in the first $2\,s$ ($t=0$ to $t=2$): $S_1 = \frac{1}{2}g(2)^2 = 2g$.Distance in the second $2\,s$ ($t=2$ to $t=4$): $S_2 = \frac{1}{2}g(4)^2 - \frac{1}{2}g(2)^2 = 8g - 2g = 6g$.Distance in the third $2\,s$ ($t=4$ to $t=6$): $S_3 = \frac{1}{2}g(6)^2 - \frac{1}{2}g(4)^2 = 18g - 8g = 10g$.The ratio is $S_1 : S_2 : S_3 = 2g : 6g : 10g = 1 : 3 : 5$.

$A$ body falls from a height $h = 200\,m$ (at New Delhi). The ratio of distance travelled in each $2\,s$ interval during $t = 0$ to $t = 6\,s$ of the journey is

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