Ratio of distances covered by object falling freely under gravity in 1

1.	Ratio of distances covered by object falling freely under gravity in 1 ^(st),2^(nd)&3^(rd)second 18 ........
Answer» `1:3:5` `1:2:5` `1:4:9` `1:5:9` Solution :Distance covered in `n^(TH)`second, In `d _(n) = v _(0)t + a/2 (2n -1) , v_(0) =0,` `d_(n) = (a)/(2) (2n -1)` `THEREFORE ` For `n =1, d _(1) = g /2 (2XX 1-1) = g/2` For `n = 2, d_(2) = g/2 (2 xx 2-1) = (3g)/(2)` For `n =3, d _(3) = g/2) (2 xx 3-1) = (5G)/(2)` `therefore d _(1) : d_(2) :d _(3) = (g)/(2) : (3g)/(2) : (5g)/(2) = 1:3:5`

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