A freely falling body travels $\_\_\_\_\_\_$ percentage of total distance in 5th second

Distance fell in first 5 second is s = 1/2 g(5²)​ = 25g/2​

Distance fell in first 4 second is s_0​ = 1/2 g(4²)​ = 16g/2

so distance covered in 5th second will be s−s₀​ = 9g/2​

so a percentage of total distance covered will be 

$\frac{9g/2}{25g/2}$ × 100 = 36%