Question

distance travelled by a body falling freely starting from rest in the 1st,2nd and 3rd seconds are in the ratio?​

Answer 1

Explanation:

Initial velocity of the body u=0

Distance covered in tthdecond, SS =u+ 21

g(2t−1)

∴ S t =0+ 21

g(2t−1)= 21

g(2t−1)

Distance travelled in first second i.e. t=1, S 1

= 21

g(2×1−1)= 21g

Distance travelled in 2nd second i.e. t=2, S 2

= 21

g(2×2−1)= 23g

Distance travelled in third second i.e. t=3, S = 21

g(2×3−1)=25 g

⟹ S 1 :S 2:S 3

=1:3:5