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The ratio of the distance traveled by a freely falling body in the $1^{st}$, $2^{nd}$, $3^{rd}$ and $4^{th}$ second :
A
$1:1:1:1$
B
$1:2:3:4$
C
$1:4:9:16$
D
$1:3:5:7$
Detailed Solution
Distance in the $n^{th}$ second: $S_n = u + \frac{a}{2}(2n - 1)$
For free fall from rest, u = 0: $S_n = \frac{g}{2}(2n - 1) \propto (2n - 1)$
$S_1 : S_2 : S_3 : S_4 = [2(1)-1] : [2(2)-1] : [2(3)-1] : [2(4)-1]$
$= 1 : 3 : 5 : 7$