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 of the following graphs correctly represents the time variation of Δx=(x2−x1), the relative position of the second stone with respect to the first till 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 Up to the first 6 seconds, the positions of the two stones are given by x1=u1t−12gt2 and x2=u2t−12gt2 ∴Δx=x2−x1=(u2−u1)t Since Δx varies linearly with t, the graph is a straight line till t = 6 s. From t = 6 s to t = 10 s, x1=0 and x2=u2t−12gt2 ∴Δx=x2−x1=u2t−12gt2 Thus (x2−x1) versus t graph is not linear; it is a parabolic curve. Now at t = 10 s, Δx=0 ∴0=10u2−12×g×(10)2 ⇒u2=5g=50 ms−1 Hence, from t = 6 s to t = 10 s, Δx=50t−5t2. Hence, the correct choice is (a).

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