Question

what will be the ratio of the distance moved by a freely falling body from rest in 4th and 5th second of journey?

Answer 1

2:3 OK please do it yourself answer is thia

Answer 2

using the eq of motion:

s = ut + 1/2gt^2... since its from rest, u=0, thus

s = 1/2gt^2

Now the distance moved during the fourth second will be from t= 3 sec to t = 4 sec

= 1/2g(4)^2 - 1/2g(3)^2

= 1/2g(7)

=7/2 g

Similarly, distance moved during fifth second will be from t = 4 sec to t = 5sec

= 1/2g(5)^2 - 1/2g(4)^2

= 1/2g(9)

= 9/2 g

Thus the ratio would be 7 : 9 or 9: 7

hope it helps