Новое решение

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

правка #20536 позже →

en/2.2.13.md +22 −0

		@@ -0,0 +1,22 @@
		+	### Statement
		+
		+	$2.2.13.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	### Statement
		+
		+	$2.2.13.$ An artilleryman fires a cannonball of mass $m$ from a cannon so that it falls into the enemy camp. Baron Munchausen, whose mass is $5m$, sits on the cannonball flying out of the cannon. What fraction of the way to the enemy camp will he have to walk?
		+
		+	### Solution
		+
		+	The cannon transfers a specific impulse to the projectile such that a projectile of mass $m$ reaches the enemy camp. From kinematics, it is known that the range of a body thrown at an angle to the horizontal is proportional to the square of its initial velocity ($S \propto v^2$).
		+
		+	The mass of the projectile together with Munchausen is $6m$. Since the same impulse is transferred, the initial velocity will be six times less ($v' = v/6$). The distance covered by such a projectile will be $1/36$ of the original distance to the camp. This means that the remaining $35/36$ of the way will have to be walked.
		+
		+	#### Answer
		+	$S/L = 35/36$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

	@@ -0,0 +1,22 @@
			### Statement

			$2.2.13.$ [Insert the problem statement]

			### Solution

			### Statement

			$2.2.13.$ An artilleryman fires a cannonball of mass $m$ from a cannon so that it falls into the enemy camp. Baron Munchausen, whose mass is $5m$, sits on the cannonball flying out of the cannon. What fraction of the way to the enemy camp will he have to walk?

			### Solution

			The cannon transfers a specific impulse to the projectile such that a projectile of mass $m$ reaches the enemy camp. From kinematics, it is known that the range of a body thrown at an angle to the horizontal is proportional to the square of its initial velocity ($S \propto v^2$).

			The mass of the projectile together with Munchausen is $6m$. Since the same impulse is transferred, the initial velocity will be six times less ($v' = v/6$). The distance covered by such a projectile will be $1/36$ of the original distance to the camp. This means that the remaining $35/36$ of the way will have to be walked.

			#### Answer
			$S/L = 35/36$

			#### Answer

			[Insert a concise answer or boxed result]