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Question

A ball is projected horizontally from top of a 80 m deep well with velocity 10 m/s. Then particle will fall on the bottom at a distance of (all the collisions with the wall are elastic)

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A
5 m from A
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B
5 m from B
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C
2 m from A
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D
2 m from B
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Solution

The correct options are
B 5 m from B
C 2 m from A
Total time taken by the ball to reach at bottom =2Hg=2×8010=4 sec
Let time taken in one collision is tcollision. Then tcollision×10=7 since horizontal velocity is 10 m/s
tcollision=0.7 sec.
No. of collisions =4.7=557 (5th collisions from wall B)
Horizontal distance traveled in between 2 successive collisions =7m
Horizontal distance traveled in 5/7 part of collisions = The 1st collision will be with wall B, 2nd collision with wall A, 3rd collision with wall B, 4th collision with wall A and 5th collision will be with wall B. After 5th collision, the ball will cover a distance of 57×7=5m from wall B i.e it will have collision at a distance of 2m from wall A.

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