Bullet loss half of the its velocity in passing through a plank what is the least number of bank required to stop the bullet

							
Let   a = acceleration/deceleration of the bullet inside the plank.
let  s = thickness of the plank
let the initial velocity of the bullet before hitting the plank: = u
final velocity = v = u - u/n = u (n-1)/n
   v²  = u² + 2 a s
   u² (n-1)² / n² = u² + 2 a s
   2 a s = u² [ (n-1)² - n² ] / n²    =  - (2n - 1) u² / n²
   a = - (2 n - 1) u² / (n² 2 s)
let S be the thickness of the N planks kept one after another to stop the bullet.  The deceleration is the same.  The initial velocity is the same.
   v² = 0 = u² - 2 S * (2 n - 1) u² / (n² 2 s) 
           1 = S/s * (2 n  -1) / n²
  
    N=   S/s = n² / (2 n - 1)  =  1/2 * [ (2n² -n) + n ] / (2n -1)
               = 1/2 *  [ n + n / (2n -1) ]
               = n/2  +  n/(4n-2) 
               the second term above is between 0 and 1.
 N =  number of planks =    [ n/2 ] + 1
           where [ n/2 ]  is the greatest integer function for n/2.
 
now in this question
 
n = 2
 
hence
 
total number of plank required = [2/2] + 1
 
 = 1 + 1
 = 2