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+### Statement
+
+$2.2.12.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+$2.2.12.$ A projectile explodes at the highest point of its trajectory at a horizontal distance $L$ from the cannon into two identical fragments. One of them returned to the cannon along the original trajectory of the projectile. Where did the second fragment fall?
+
+### Solution
+
+The projectile explodes at the highest point of the trajectory, which means it has no vertical velocity component. Let its velocity be $\vec{v}$. Since one of the fragments returned to the cannon, its velocity will be $-\vec{v}$.
+
+Let's write the law of conservation of momentum on the horizontal axis:
+$$m_0 \vec{v} = -m \vec{v} + m \vec{v}_2$$
+$$\vec{v}_2 = \frac{m_0 + m}{m} \vec{v} = 3\vec{v}$$
+
+The velocity of the second fragment is $3$ times greater than $v$, which means it will cover a horizontal distance to the ground of $3L$ from the point of the explosion, or $4L$ from the cannon.
+
+#### Answer
+At a horizontal distance of $4L$ from the cannon.
+
+#### Answer
+
+[Insert a concise answer or boxed result]