Question

a body is dropped from a height of 100 meters. ratio of distance covered after 1 second,2 second , 3 second of it's downward journey is​

Answer 1

1:3:5

Initial velocity of the body u=0

Distance covered in tth second, St=u+21g(2t−1)

∴ St=0+21g(2t−1)=21g(2t−1)

Distance travelled in first second i.e. t=1, S1=21g(2×1−1)=21g

Distance travelled in 2nd second i.e. t=2, S2=21g(2×2−1)=23g

Distance travelled in third second i.e. t=3, S3=21g(2×3−1)=25g

⟹ S1:S2:S3=1:3:5

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