Two diamonds begin a free fall from rest from the same height 0.9 s apart. How long after the first object begins to fall will the two objects be 13 m apart

Explanation:

Suppose the first diamond falls for t seconds

The downward displacement S₁ = u t + g t²

here u = 0 , because it was at rest initially

Thus S₁ = g t²              I

Similarly second diamond moves for time ( t - 0.9 ) sec

The distance covered by it is

S₂ = g (t - 0.9 )²

The separation between the two is 13 m

Thus S₁ - S₂ = 13 m

g t² - g t² + 0.9 g t - 0.4 g = 13

or  9 t = 13 - 4 = 9

Hence t = 1 sec

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer