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Question

# Two balls are projected upward simultaneously with speeds 40 m/s and 60 m/s. Relative position (x) of second ball w.r.t. first ball at time t=5 s is : [Neglect air resistance]

A
120 m
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B
80 m
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C
20 m
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D
100 m
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Solution

## The correct option is D 100 mLet the velocity of projections be v1 and v2. Relative velocity of ball 2 w.r.t 1, urel=v2−v1=60−40=20 m/s Since, both the balls are moving under constant acceleration due to gravity. Therefore, relative acceleration of ball 2 w.r.t ball 1 will be zero. ⇒arel=0 Position of ball 2 w.r.t ball 1, srel=urelt+12arelt2 ⇒srel=20×5 ⇒srel=100 m

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