Новое решение

Valter правка от 30 авг. 2026 г., 19:16

правка #20428 позже →

en/2.3.4.md +27 −0

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		+	### Statement
		+
		+	$2.3.4.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	### Statement
		+
		+	$2.3.4.$ Estimate the average force developed by a person's legs when landing after jumping from a second-floor window.
		+
		+	### Solution
		+
		+	Let's use standard physical estimates for a human. Let the mass of the person be $m = 68\text{ kg}$. The height of a second-floor window is approximately $H = 5\text{ m}$. When landing, a person bends their knees to cushion the impact. Let the stopping distance (the vertical displacement of the center of mass during the squat) be $h = 0.68\text{ m}$.
		+
		+	According to the work-energy theorem, the work done by the average force of the legs goes into bringing the person's kinetic energy to zero, which originally came from the potential energy drop:
		+	$$F \cdot h \approx mgH$$
		+	$$F = mg\frac{H}{h}$$
		+
		+	Substituting the estimated values:
		+	$$F = 68\text{ kg} \cdot 10\text{ }\frac{\text{m}}{\text{s}^2} \cdot \frac{5\text{ m}}{0.68\text{ m}} = 5000\text{ N} = 5\text{ kN}$$
		+
		+	#### Answer
		+	$F \approx 5\text{ kN}$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$2.3.4.$ [Insert the problem statement]

			### Solution

			### Statement

			$2.3.4.$ Estimate the average force developed by a person's legs when landing after jumping from a second-floor window.

			### Solution

			Let's use standard physical estimates for a human. Let the mass of the person be $m = 68\text{ kg}$. The height of a second-floor window is approximately $H = 5\text{ m}$. When landing, a person bends their knees to cushion the impact. Let the stopping distance (the vertical displacement of the center of mass during the squat) be $h = 0.68\text{ m}$.

			According to the work-energy theorem, the work done by the average force of the legs goes into bringing the person's kinetic energy to zero, which originally came from the potential energy drop:
			$$F \cdot h \approx mgH$$
			$$F = mg\frac{H}{h}$$

			Substituting the estimated values:
			$$F = 68\text{ kg} \cdot 10\text{ }\frac{\text{m}}{\text{s}^2} \cdot \frac{5\text{ m}}{0.68\text{ m}} = 5000\text{ N} = 5\text{ kN}$$

			#### Answer
			$F \approx 5\text{ kN}$

			#### Answer

			[Insert a concise answer or boxed result]