Push force in a vertical jump

Problem

A person of $70\;\text{kg}$ jumps vertically, reaching a height of $40\;\text{cm}$. Estimate the average force (above their weight) that their feet exert on the ground during the push-off phase, assuming their centre of gravity lowers by about $30\;\text{cm}$ while crouching before the jump. Use the CM theorem.

Solution

Velocity at take-off. To rise by $h = 0{,}40\;\text{m}$, at the moment of leaving the ground one needs $v = \sqrt{2 g h} = \sqrt{2\cdot 9{,}81\cdot 0{,}40} \approx 2{,}8\;\text{m/s}$

Acceleration during the push-off (over a distance $d = 0{,}30\;\text{m}$, starting from rest): $a = \frac{v^2}{2 d} = \frac{2{,}8^2}{2\cdot 0{,}30} \approx 13\;\text{m/s}^2$

Net force on the CM (CM theorem, $\sum F_\text{ext} = M a_\text{CM}$). The net upward force is $N - Mg = M a$, so the part of $N$ beyond the weight is $F_\text{netta} = M a = 70\cdot 13 \approx 0{,}9\cdot 10^3 \sim 10^{3}\;\text{N}$

Total force on the ground: $\ev{F_\text{tot} = M(g + a) = 70\cdot(9{,}81 + 13) \approx 1{,}5\cdot 10^{3}\;\text{N}}$ roughly twice the body weight.

Links

Topics: Centre of mass Concepts: Centre of mass · Newton’s second law Skills: Fermi estimates