The bouncy problem

As $\triangle ABD$ is $30°-60°-90°$ $BD=\frac{1}{2}AB=1$ cm $\therefore EN=12$ . And by pythagorean theorem $AD=\sqrt{3}$ .

Then $\triangle NAE$ also is $30°-60°-90°\Rightarrow \triangle NAE \sim \triangle ABD\Rightarrow \frac{EN}{AD}=\frac{AN}{2}$ $\Rightarrow \frac{12}{3}=\frac{AN}{2} \therefore AN=\frac{24}{\sqrt{3}}=8\sqrt{3}$

Last, applying pythagorean theorem in $\triangle ABN$ , $AN=\sqrt{(8\sqrt{3})^2+2^2}=\boxed{14 cm}$

My way I a kind of dumb so

I extend the shape into a big right angled triangle by adding a small triangle on the corner. And I will call the extra corner D. The corner ABD is 180-120<360-90-90-60>=60degrees. The corner BAD is 180-90=90degrees. And the ADB is 180-90-60=30 degrees. Now I can get the length of BD by doing this procedure 2/sin30=BD/sin90(1) which is 4. Together, the length of DC is 11+4=15 cm. then I use sin30[15/sin60]to get NC which turns out to be 8.66. And finally I use the Pythagorean theorem th find BN: 8.66^2+11^2=14^2

my solution ...ANOTHER solution