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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 sec
Considering 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=uBuA=(40)(80)=120 m/s
Acceleration of B w.r.t A
aBA=aBaA=(g)(g)=0

Using second equation of motion:
SBA=uBAt+12aBAt2
160=120t
t=1.33 s

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