Do Some Physics Before Shooting!

Let's do this short. If the thickness of one book is $s$ , then thickness of $n$ books is $ns$ .

Now we would implement proportionality concept. We know $s=\frac{v^2-u^2}{2a}$ As the deceleration, a is constant, we can write $s\propto v^2-u^2$ $\rightarrow \frac{s_2}{s_1}=\frac{{v_2}^2-{u_2}^2}{{v_1}^2-{u_1}^2}$ For the 1st (couple's) case, $s_1=s, u_1=457.2m/s, v_1=400m/s$ And for the 2nd (and safe) case, $s_2=ns, u_2=457.2m/s, v_2=0m/s \text{(as the bullet must stop)}$

Finally, $\frac{ns}{s}=\frac{0-457.2^2}{400^2-457.2^2}$ $\rightarrow n=\lceil 4.26 \rceil = 5 (ans)$ As we can't take 4.26 books, we need at least five books to stop the bullet.

Using formula $\upsilon^{2}=\upsilon_{0}^{2}-2as$ where:

$a$ is (negative) acceleration of the bullet caused by thickness of book,

$s$ is distance traveled (in this case thickness of one book),

$\upsilon_{0}$ is starting speed of the bullet and

$\upsilon_{s}$ is speed of the bullet after distance traveled $s$ ,

we get $a=\dfrac{\upsilon_{0}^{2}-\upsilon_{s}^{2}}{2s}$ .

If $ns$ is thickness of $n$ books, then $ns=\dfrac{\upsilon_{0}^{2}-\upsilon_{ns}^{2}}{2a}$ .

Substituting $\upsilon_{ns}$ with $0$ and $a$ with $\dfrac{\upsilon_{0}^{2}-\upsilon_{s}^{2}}{2s}$ :

$ns=\dfrac{\upsilon_{0}^{2}}{2\frac{\upsilon_{0}^{2}-\upsilon_{s}^{2}}{2s}}$

$\Rightarrow n=\dfrac{\upsilon_{0}^{2}}{\upsilon_{0}^{2}-\upsilon_{s}^{2}}=4.27914...$

In order to be sure that bullet will stop before reaching him, he should stack up at least $\boxed{5}$ books.