Question

# Two bodies A and B are thrown simultaneously. A is projected vertically upwards with 80 m/s speed from the ground and B is projected vertically downwards from a height of 160 m with 40 m/s and along the same line of motion. The time taken for the two bodies to collide is

A
1 sec
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B
1.33 sec
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C
1.66 sec
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D
2 sec
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Solution

## The correct option is B 1.33 secConsidering the ground to be origin and upward direction as positive, Initial position of A (xi)A=0 m Initial position of B (xi)B=160 m Now, considering motion w.r.t A, Initial position of B w.r.t A (xi)BA=160 m Final position of B w.r.t A (xf)BA=0 m (Since both collide and are at same location) Initial velocity of B w.r.t A uBA=uB−uA=(−40)−(80)=−120 m/s Acceleration of B w.r.t A aBA=aB−aA=(−g)−(−g)=0 Using second equation of motion: SBA=uBAt+12aBAt2 ⟹−160=−120t t=1.33 s

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