Новое решение

Valter правка от 3 сент. 2026 г., 18:52

правка #20534 позже →

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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]

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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]