Question

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

Answer 1

The time taken is 1 second

Explanation:

Suppose the first diamond falls for t seconds

The downward displacement S₁ = u t + $\frac{1}{2}$ g t²

here u = 0 , because it was at rest initially

Thus S₁ = $\frac{1}{2}$ g t²              I

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

The distance covered by it is

S₂ = $\frac{1}{2}$ g (t - 0.9 )²

The separation between the two is 13 m

Thus S₁ - S₂ = 13 m

$\frac{1}{2}$ g t² - $\frac{1}{2}$ g t² + 0.9 g t - 0.4 g = 13

or  9 t = 13 - 4 = 9

Hence t = 1 sec