Question

A body falling from rest describes distance s1 , s2 and s3 in the first, second and third seconds of its fall. Then the ratio of s1 : s2 : s3 is

Answer 1

Answer:

body is falling from rest describes distances S1 S2 and S3 in the first second and third seconds of its fall. ... Let u be the initial velocity of the body . Theng S = u t + 2 and S 1 + S 2 + S 3 = 1 2 × g × 3 2 = 4 .

Answer 2

Answer:

S=ut+12gt2S=ut+12gt2, u=0u=0

S=12gt2S=12gt2

Here s is displacement.

Distance dropped in 1st1st second, S1=12g×(1)2=12g×1S1=12g×(1)2=12g×1

Distance dropped in 2nd2nd second, S2=12g×(2)2−12g×(1)2=12g×3S2=12g×(2)2−12g×(1)2=12g×3

Distance dropped in 3rd3rd second, S3=12g×(3)2−12g×(2)2=12g×5S3=12g×(3)2−12g×(2)2=12g×5

S1:S2:S3=1:3:5