Explanation:
Suppose the first diamond falls for t seconds
The downward displacement S₁ = u t + [tex]\frac{1}{2}[/tex] g t²
here u = 0 , because it was at rest initially
Thus S₁ = [tex]\frac{1}{2}[/tex] g t² I
Similarly second diamond moves for time ( t - 0.9 ) sec
The distance covered by it is
S₂ = [tex]\frac{1}{2}[/tex] g (t - 0.9 )²
The separation between the two is 13 m
Thus S₁ - S₂ = 13 m
[tex]\frac{1}{2}[/tex] g t² - [tex]\frac{1}{2}[/tex] g t² + 0.9 g t - 0.4 g = 13
or 9 t = 13 - 4 = 9
Hence t = 1 sec