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

Question

A body starts to fall freely under gravity. The distances covered by it in first, second and third seconds are in ratio (A)  1:3:5  (B)  1:2:3  (C)  1:4:9  (D)  1:5:6

Tardigrade Admin · Accepted Answer

According to equation of motion, distance covered in nth sec. Sn =u=(a/2)(2 n-1) Sn =(a/2)(2 n-1) ∴ S1: S2: S3= 2(1)-1: 2(2)-1: 2(3)-1 =1: 3: 5

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

$1 : 3 : 5$

$1 : 2 : 3$

$1 : 4 : 9$

$1 : 5 : 6$

According to equation of motion, distance covered in $n$ th $sec$ .
$S_{n} = u = \frac{a}{2} (2 n - 1)$
$S_{n} = \frac{a}{2} (2 n - 1)$
$∴ S_{1} : S_{2} : S_{3} = {2 (1) - 1} : {2 (2) - 1} : {2 (3) - 1}$
$= 1 : 3 : 5$

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

1:3:5

1:2:3

1:4:9

1:5:6

According to equation of motion, distance covered in nth sec. Sn​=u=2a​(2n−1) Sn​=2a​(2n−1) ∴S1​:S2​:S3​={2(1)−1}:{2(2)−1}:{2(3)−1} =1:3:5

$1 : 3 : 5$

$1 : 2 : 3$

$1 : 4 : 9$

$1 : 5 : 6$

According to equation of motion, distance covered in $n$ th $sec$ .
$S_{n} = u = \frac{a}{2} (2 n - 1)$
$S_{n} = \frac{a}{2} (2 n - 1)$
$∴ S_{1} : S_{2} : S_{3} = {2 (1) - 1} : {2 (2) - 1} : {2 (3) - 1}$
$= 1 : 3 : 5$