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