Question

Two diamonds begin a free-fall from the rest from the same height ,1.0 s apart. How long after the first diamond begins to fall will the two diamonds be 10 m apart

Answer 1

Refer Figure .
In this figure 'h' is the displacement of diamond A which is dropped 1 second after the diamond B just at the instant when Diamond B has displaced by 'H' and the separation between them is 10 meters .
=> H-h = 10 ......... (1)
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Now ;
H = ½×gt² = 5t² ........... (2)
where t is the time taken by B to cover H height . and g is acceleration due to gravity.
then the time taken by A to reach at h is given by = (t-1) seconds . (obviously)
then ;
h= ½×g×(t-1)² = 5(t-1)² ...... (3)
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Using (1) ,(2),(3) we get ;
5t² - 5(t-1)² = 10
=> t² - (t-1)² = 2
=> t = 3/2 seconds .
This is the time when B has fallen down by H (our required time )
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hope it helps !

Answer 2

Answer:

In ‘t’ second the first body falls through a distance,

S1 = 0 + ½ gt2 = 5t2

And the second body falls through a distance,

S2 = 0 + ½ g(t – 1)2

=> S2 = ½ gt2 – gt + ½ g = 5t2 +5 -10t

And,

S1 – S2 = 10

=> ½ gt2 – (½ g – gt + ½ g) = 10

=>  5t2 -5t2 - 5 +10t = 10

=> t = 15 /10

 = 1.5 sec