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)

• A. $5$ m from A
• B. $5$ m from B
• C. $2$ m from A
• D. $2$ m from B

Answer 1

Answer: (B), (C)

The correct options are
B $5$ m from B
C $2$ m from A

Total time taken by the ball to reach at bottom $=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2\times 80}{10}}=4sec$
Let time taken in one collision is $t_{collision}$. Then $t_{collision}\times 10=7$ since horizontal velocity is 10 m/s
$\Rightarrow t_{collision}=0.7$ sec.
No. of collisions $=\frac{4}{.7}=5\frac{5}{7}$ (5th collisions from wall B)
Horizontal distance traveled in between $2$ successive collisions $=7m$
$\therefore$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 $\frac{5}{7}\times 7=5m$ from wall B i.e it will have collision at a distance of 2m from wall A.