The given problem calculates the maximum height $h$ from which a person can jump without breaking their lower leg bones, using physical parameters such as the bone's fracture strength, cross-sectional area, and the person's mass.

Let’s break this down step by step:

Problem Data

1. Mass of the person: $m = 75 \, \text{kg}$
2. Length of deceleration path: $d = 0.6 \, \text{m}$
3. Bone fracture strength: $\sigma = 170 \cdot 10^6 \, \text{N/m}^2$
4. Cross-sectional area of bone: $A = 2.5 \cdot 10^{-4} \, \text{m}^2$
5. Control result: $h = 68 \, \text{m}$

Steps to Solve

1. Force that the bones can withstand: The maximum force the bone can resist is: $F_{\text{max}} = \sigma \cdot A$ Substituting the given values: $F_{\text{max}} = 170 \cdot 10^6 \cdot 2.5 \cdot 10^{-4}$ $F_{\text{max}} = 42,500 \, \text{N}$

2. Velocity of the person just before landing: Using conservation of energy (potential to kinetic energy): $m \cdot g \cdot h = \frac{1}{2} m \cdot v^2$ Simplifying for $v$, we get: $v = \sqrt{2 g h}$ where $g = 9.81 \, \text{m/s}^2$.

3. Force during landing (impulse consideration): The person decelerates to rest over a distance $d$. Using work-energy theorem: $F_{\text{avg}} \cdot d = \frac{1}{2} m \cdot v^2$ Solving for $F_{\text{avg}}$: $F_{\text{avg}} = \frac{\frac{1}{2} m v^2}{d}$ Since $v^2 = 2 g h$ from earlier: $F_{\text{avg}} = \frac{\frac{1}{2} m (2 g h)}{d} = \frac{m \cdot g \cdot h}{d}$

4. Set the force equal to $F_{\text{max}}$: To avoid breaking bones: $F_{\text{avg}} \leq F_{\text{max}}$ Substituting for $F_{\text{avg}}$: $\frac{m \cdot g \cdot h}{d} = F_{\text{max}}$ Solving for $h$: $h = \frac{F_{\text{max}} \cdot d}{m \cdot g}$

Substitute the values:

• $F_{\text{max}} = 42,500 \, \text{N}$
• $d = 0.6 \, \text{m}$
• $m = 75 \, \text{kg}$
• $g = 9.81 \, \text{m/s}^2$

$h = \frac{42,500 \cdot 0.6}{75 \cdot 9.81}$

$h = \frac{25,500}{735.75} \approx 34.7 \, \text{m}$

Explanation of Discrepancy:

The control result states $h = 68 \, \text{m}$, which may arise if additional assumptions (e.g., time of impact or force distribution) are considered in a more complex setup. Let me know if you'd like further clarification or another detailed derivation considering impulse.

Follow-Up Questions:

1. How is energy conservation related to this problem?
2. Why is impulse crucial in determining bone safety upon landing?
3. How would increasing the deceleration path $d$ affect the height?
4. How does a change in mass $m$ influence the result?
5. Can we include air resistance to refine the calculation?

Tip: When calculating forces during impacts, consider both the deceleration path and time to distribute the forces safely!