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Question

# Two stones are thrown up simultaneously with initial speeds of u1 and u2 (u2>u1). They hit the ground after 6 s and 10 s respectively. Which graph in figure correctly represents the time variation of Δx=(x2−x1), the relative position of the second stone with respect to the first upto t=10 s? Assume that the stones do not rebound after hitting the ground.

A
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B
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C
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D
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Solution

## The correct option is A Upto 6 s relative velocity of stone 2 w.r.t. 1 is urel=u2−u1 arel=g−g=0 (take downward direction as positive) Using second equation of motion, we get Srel=urelt+12arelt2 Δx=(u2−u1)×t Initially relative acceleration between them is zero, so distance between them will increase linearly. Hence, Δx increases linearly with t upto 6 s From t=6 s to t=10 s, first stone is at rest arel=g Δx=(u2−u1)t−12gt2 (in the downward journey Δx decreases parabolically) Hence, the correct answer is option (a)

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