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(C) For a body falling from rest,the distance traveled in time $t$ is given by $s = \frac{1}{2} g t^2$.The distance traveled in the first interval ($t

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$1 : 3 : 5$

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(C) For a body falling from rest,the distance traveled in time $t$ is given by $s = \frac{1}{2} g t^2$.The distance traveled in the first interval ($t = 0$ to $t = t_0$) is $s_1 = \frac{1}{2} g t_0^2$.The distance traveled in the first $2 t_0$ is $s_{0-2t_0} = \frac{1}{2} g (2 t_0)^2 = 4 \left( \frac{1}{2} g t_0^2 \right) = 4 s_1$.Thus,the distance traveled in the second interval ($t = t_0$ to $t = 2 t_0$) is $s_2 = s_{0-2t_0} - s_1 = 4 s_1 - s_1 = 3 s_1$.The distance traveled in the first $3 t_0$ is $s_{0-3t_0} = \frac{1}{2} g (3 t_0)^2 = 9 \left( \frac{1}{2} g t_0^2 \right) = 9 s_1$.Thus,the distance traveled in the third interval ($t = 2 t_0$ to $t = 3 t_0$) is $s_3 = s_{0-3t_0} - s_{0-2t_0} = 9 s_1 - 4 s_1 = 5 s_1$.The ratio of distances traveled in each interval is $s_1 : s_2 : s_3 = s_1 : 3 s_1 : 5 s_1 = 1 : 3 : 5$.

$A$ body falls from a large height. The ratio of distance traveled in each time interval $t_0$ during $t = 0$ to $t = 3\, t_0$ of the journey is

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