Question

the ratio's of the distance covered by a freely falling practice realesed from rest in first second ........... nth second of the motion ​

Answer 1

Answer:

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

pls mark me as brainliest

pls follow me