Let a person of mass ( m ) = 10kg. then his weight will be 98N. Now, when he jumps from 1m height on the ground, the impact force becomes 9800N ( considering deflection of ground after impact is 1cm ). 100 times. But i am highly confused. 100 times force should simply crush a human! how the force is neutralized?

for calculating impact force: http://hyperphysics.phy-astr.gsu.edu/hbase/flobi.html


Well for the human body, to calculate the impact force is not as simple as you think, because we are soft objects and have a large cross section area (of course depends on which part of your body you land). The involved factors are:

  • Velocity upon impact: with initial speed $v_0=0$ and s the distance of the fall, then $$v=2gs$$

  • Rate of deceleration: with $d$: deceleration distance: $$a=\frac{v^2}{2d}$$

  • G-force: with g: gravitational acceleration: $$G = \frac{a}{g} $$

  • Force of impact: W: weight $$F=WG$$

  • duration of fall (redundant here): $$t=\sqrt{\frac{2s}{g}}$$

  • Impact pressure: A: cross-section of impact $$P = \frac{F}{A} $$

  • Force withstand based on stress absorption (specific to object denoted $\sigma$):$$F_{abs}=\sigma A$$

Numerical estimate for your example:

Simplification: to simplify the deal with force withstand of your body, we can consider landing on a surface that deflects $0.3 m$ upon impact, so:

  • $$v=\sqrt{2*9.8*1}\approx 4.43 m/s$$

  • $$a=\frac{4.43^2}{2*0.3}\approx 32.6 m/s^2$$

  • $$G=\frac{32.6}{9.8}\approx 3.33$$

  • $$F=10*9.8*3.33=326.34 N$$

Much lower than the impact force you had in mind! Now if you fall on your hip, the impact pressure is (considering a hip of $A=0.025 m^2$): $$P=\frac{326.34}{0.025}=13053 Pa$$ Note we were working with $m=10 kg$

  • $\begingroup$ So is it like, the object fallen is getting a force 100 times than it's weight and absorbing it somehow ( factors you've mentioned ) ? $\endgroup$ – hello16 Aug 4 '14 at 12:24
  • $\begingroup$ I added a numerical example for you. $\endgroup$ – Phonon Aug 4 '14 at 12:35

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