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<blockquote data-quote="UngeheuerLich" data-source="post: 9364721" data-attributes="member: 59057">Which is wrong... v is proportional to falling time. Not falling distance:Consider:Energy conservation gives us:(I) E(end) = E(beginning)(II) E(beginning) = E(pot) = m*g*h(III) E(end) = E(kin) = 1/2*m*v^2&nbsp; &nbsp; (if you neglect energy lost due to friction)so II and III in I:1/2*m*v^2 = m*g*hso after some simple transformations you getv = squareroot(2*g*h)So the speed is proportional to the square root of the fallen distance (if you neglect friction... otherwise it is even lower).Now the question is, if kinetic energy or momentum deals damage. And a short google search is that experiments with a falling ball in clay show that the damage is proportional to speed^2. Which is proportional to the energy transferred. And this is proportional to distance fallen. So linear increasing damage makes sense.</blockquote>

[QUOTE="UngeheuerLich, post: 9364721, member: 59057"] Which is wrong... v is proportional to falling time. Not falling distance: Consider: Energy conservation gives us: (I) E(end) = E(beginning) (II) E(beginning) = E(pot) = m*g*h (III) E(end) = E(kin) = 1/2*m*v^2 (if you neglect energy lost due to friction) so II and III in I: 1/2*m*v^2 = m*g*h so after some simple transformations you get v = squareroot(2*g*h) So the speed is proportional to the square root of the fallen distance (if you neglect friction... otherwise it is even lower). Now the question is, if kinetic energy or momentum deals damage. And a short google search is that experiments with a falling ball in clay show that the damage is proportional to speed^2. Which is proportional to the energy transferred. And this is proportional to distance fallen. So linear increasing damage makes sense. [/QUOTE]

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