Two objects are dropped from the same point after an interval of 2 seconds. Vertical distance between them 4 seconds after the release of the second object will be (g=10 m/ s2)

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Solution

You can use Newton's second equation of motion here.
The first ball has traveled for 6 seconds under gravity with acceleration g and initial velocity=0.
The second ball has traveled for 4 seconds under gravity with acceleration g and initial velocity=0.
Distance between the 2 balls at the end of 4 seconds after the first ball is dropped = 1/2g(t1^-t2^2) = 1/2*10(36-16)
= 100 m

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