A bullet loses `1//20^(th)` of its velocity in passing through a plank. Find the least numbers of such planks required just to stop the bullet.

Explanation

`{((19v)/(20))^2 = v^2 - 2a xx x implies ((39)/(400)) v^2 = 2ax .....(i)}`
Say n be the number of planks required to stop the bullet
`therefore 0^2 = v^2 – 2a xx nx`
`implies v^2 = 2a xx nx `
Putting (ii) in (i), we get
`((39)/(400)) xx 2a xx nx = 2 ax`
`n = (400)/(39) = 10.25`
So, the bullet will stop inside the `11^(th)` plank.