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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 and wall is smooth)
(Take g=10 m/s2)

A
5 m from A
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B
4 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 option is C 2 m from A
As the velocity initially in y-direction =0
Total time taken by the ball to reach at the bottom =2Hg=2×8010=4 sec
Let time taken in one collision be t,
Then t×10=7
t=0.7sec
No of collisions =40.7=557 (5th collisions from wall B)
Horizontal distance travelled in between 2 successive collisions =7m
Horizontal distance travelled in 57 part of collisions =57×7=5 m
Hence, distance from B when it reaches the ground =5 m
Similarly, Distance from A where it reaches the bottom =75=2 m

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