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

Solution :
As distance travelled in n^th $$\sec$$  is given by
$${s_n} = u + \frac{1}{2}a\left( {2n - 1} \right)$$
Here, $$u = 0,$$  acceleration due to gravity
$$\eqalign{ & a = 9.8\,m/{s^2} \cr & \therefore {\text{For}}\,\,{4^{th}}s,\,{s_4} = \frac{1}{2} \times 9.8\left( {2 \times 4 - 1} \right) \cr} $$
and for $${5^{th{\text{ }}}}s,\,\,{s_5} = \frac{1}{2} \times 9.8\left( {2 \times 5 - 1} \right)$$
$$\therefore \frac{{{S_4}}}{{{s_5}}} = \frac{7}{9}$$