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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.

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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)

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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