A rifle bullet loses 1/20th of its velocity in passing through a plank. What will be the least number of such planks required to just stop the bullet ?

Explanation

For one plank u=u, `v=u-(u)/(20)=(19)/(20)u`
`((19v)/(20))^(2)-v^(2)=2ax…….(1)`
For n planks u=u, v=0
`0^(2)-v^(2)=2nx……(2)`
Dividing (2), (1) `n=(-v^(2))/((19)/(20)v)^(2)-v^(2)=(1)/(1-((19)/(20))^(2))`
`=(20 xx 20)/((20+19) (20-19))=(400)/(39)=10.3`
so, 11 planks are required. The bullet shall stop in 11th plank.