A rifle bullet loses 1/20 th its veloity 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)=2anx`……..(2)
Dividing (2),(1) `n=(-v^(2))/(((19)/(20))^(2)-v^(2))=(1)/(1-((19)/(20))^(2))`
`=(20xx20)/((20+19)(20-19))=(400)/(39)`=10.3
`implies` so 11 planks are required.The bullet shall stop in 11 th plank