A body starts to fall freely under gravity. The distances covered by it in first, second and third second are in ratio

Options

A. `1:4:9 `
B. `1:2:3 `
C. `1:3: 5`
D. `1 : 5 : 6`

Answer

C

Explanation

Here, u = 0
Distance covered in 1 minute is `S_(1) =1/2 g(1)^(2) = g/2`
Distance covered in 2 minute is `S_(2) g(2)^(2) = (4g)/2`
Distance covered in 3 minute is `S_(3) g(3)^(2) = (9g)/2` Therefore , distance covered in `1^(st)` minute is `D_(1) =S_(1) = g/2`
Distance covered in `2^(nd)` minute is `D_(2) =S_(2) -S_(1) = (4g)/2 -g/2 =(3g)/2`
Distance covered in `3^(nd)` minute is `D_(3) =S_(3) -S_(2) = (9g)/2 -(4g)/2 =(5g)/2`
`:. D_(1) : D_(2) : D_(3) = g/2 : (3g)/(2) : (5g)/(2) = 1:3 :5` s `D_(3) =S_(3) -S_(2) = (9g)/2 -(4g(/2 =(5g)/2`
`:. D_(1) : D_(2) : D_(3) = g/2 : (3g)/(2) : (5g)/(2) = 1:3 :5`