Jumping off the bench

Problem

You jump down from a $50$ cm high bench and land on your feet, bending your knees over $20$ cm. Estimate the average force the floor exerts on your feet using the impulse-momentum theorem. What changes if you land stiffly, braking over $\sim 2$ cm?

Solution

Mass estimate: $m \sim 70$ kg. Landing speed after falling $h = 0{,}50$ m: $v = \sqrt{2 g h} = \sqrt{2 \cdot 9{,}81 \cdot 0{,}50} \approx 3{,}1\ \mathrm{m/s}$

During the braking distance $d$ the deceleration is nearly constant, so the stopping time is $\Delta t \approx 2d/v$. The impulse-momentum theorem along the vertical gives (the floor’s force must cancel the momentum and support the weight): $\bar F \approx \frac{m v}{\Delta t} + m g = \frac{m v^2}{2 d} + m g = m g\left(1 + \frac{h}{d}\right)$

Soft landing, $d = 20$ cm $= 0{,}20$ m: $\bar F \approx 70 \cdot 9{,}81 \left(1 + \frac{0{,}50}{0{,}20}\right) \approx 687 \cdot 3{,}5 \approx 2{,}4\cdot 10^{3}\ \mathrm{N} \sim 2\cdot 10^{3}\ \mathrm{N}$

Stiff landing, $d = 2$ cm $= 0{,}02$ m: $\bar F \approx 687 \left(1 + \frac{0{,}50}{0{,}02}\right) \approx 687 \cdot 26 \approx 1{,}8\cdot 10^{4}\ \mathrm{N} \sim 2\cdot 10^{4}\ \mathrm{N}$

Bending the knees lengthens the braking distance (and time) by a factor of $\sim 10$, so the force on the feet is reduced by the same factor: this is why you bend your legs when landing.

Final result: $\ev{\bar F \sim 2\cdot 10^{3}\ \mathrm{N}\ \text{(soft)};\quad \sim 2\cdot 10^{4}\ \mathrm{N}\ \text{(stiff)}}$

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Topics: Quantità di moto e urti Concepts: Impulso · Energia cinetica Skills: Stime di Fermi