Solution:

Concept: Distance covered in the \(n^{\text{th}}\)) second of free fall from rest.
Formula: \(h_n = u + \frac{g}{2}(2n-1)\). Since \(u=0\), \(h_n = \frac{g}{2}(2n-1)\).
Distance in 4th second (\(n=4\)): \(h_4 = \frac{g}{2}(2 times 4 - 1) = \frac{7g}{2}\).
Distance in 5th second (\(n=5\)): \(h_5 = \frac{g}{2}(2 times 5 - 1) = \frac{9g}{2}\).
Ratio: \(h_4 : h_5 = \frac{7g}{2} : \frac{9g}{2} = 7:9\).