Question : Prove that the distances traversed during equal intervals of time by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning from unity [namely 1:3:5:7........].
Doubt by Saniya
Solution :
Here,
initial velocity, u=0 m/s
Distance covered by the body during nth second is given by
Sn=u+[a(2n-1)]/2
Here u=0
a=g
so
Sn=[g(2n-1)]/2
Distance covered by body during 1st second (S1)
S1=[g(2{1}-1)]/2
S1=g/2
Similarly
S2=3g/2
S3=5g/2
S4=7g/2
and so on.
Required ratio
S1:S2:S3:S4 . . . = g/2:3g/2:5g/2:7g/2 . . .
S1:S2:S3:S4. . . = 1:3:5:7. . .
Hence, distances traversed during equal intervals of time by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning from unity.