**The answer is 11 planks**

**Solution:**

Let us assume that the thickness of the plank be **s** and acceleration provided by each plank is **a**. Using equation of motion we have,

v^{2}−u^{2}=2as

As the bullet stops finally, therefore above equation can be rewritten as,

0−u^{2}=2as

Now, let the number of planks required be **n**, therefore the equation can be written as,

−u^{2}=2asn

or,

n=−u^{2}/2as ... (1)

Now, question says that the bullet loses its speed by 1/20 on passing through plank, so the final speed of the bullet when it leaves one plank is,

v=u−u/20

v=19u/20

Again using equation of motion we have,

v^{2}−u^{2}=2as

Substituting for v in above equation we get,

(19u/20)^{2}−u^{2}=2as

361u^{2}/400−u^{2}=2as

−39u^{2}/400=2as

or,

2as=−39u^{2}/400 ... (2)

substituting equation (2) in equation (1) we get,

**So, the number of planks required is 11 because 0.26 can be considered as one plank.**