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Q.
What will be the ratio of the distance moved by a freely falling body from rest in $4^{\text{th}}$ and $5^{\text{th}}$ seconds of journey ?
AIPMTAIPMT 1989Motion in a Straight Line
Solution:
Distance covered in $n^{t h}$ second is given by
$s_{n}=u+\frac{a}{2}(2 n-1)$
Given : $u = 0 , a = g$
$\therefore s_{4}=\frac{g}{2}(2 \times 4-1)=\frac{7 g}{2} $
$s_{5}=\frac{g}{2}(2 \times 5-1)=\frac{9 g}{2} $
$ \therefore \frac{s_{4}}{s_{5}}=\frac{7}{9}$