Question

A ball is projected horizontally from top of a 80 m deep well with velocity $10m/s.$ Then particle will fall on the bottom at a distance of (all the collisions with the wall are elastic and wall is smooth)
(Take $g=10m/s^2$)

• A. $5m$ from A
• B. $4m$ from B
• C. $2m$ from A
• D. $2m$ from B

Answer 1

The correct option is C $2m$ from A

As the velocity initially in y-direction $=0$
$\therefore$ Total time taken by the ball to reach at the bottom $=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2\times 80}{10}}=4\sec$
Let time taken in one collision be $t$,
Then $t\times 10=7$
$t=0.7\sec$
No of collisions $=\frac{4}{0.7}=5\frac{5}{7}$ (5th collisions from wall B)
Horizontal distance travelled in between 2 successive collisions $=7m$
$\therefore$ Horizontal distance travelled in $\frac{5}{7}$ part of collisions $=\frac{5}{7}\times 7=5m$
Hence, distance from B when it reaches the ground $=5m$
Similarly, Distance from A where it reaches the bottom $=7-5=2m$