Why falling ants will get less damage than a human? I know this is related to the mass of the object, so that an ant falling will have a little terminal velocity and won't get much damage, when compared to a human which will reach an higher velocity.
I'm trying to explain this to myself: the energy that is "converted to damage" is the final kinetic energy $\frac{1}{2}mv_f^2$, so that a human will inevitably result in more damage, because she would reach a higher velocity. But I'm wondering how the mass intervenes in this:


*

*It intervenes indirectly in the fact that having more mass (well, with respect to the section area of course) leads to a weaker air resistance and, thus, to a higher velocity

*It also intervenes "directly" because kinetic energy is higher for an higher mass, i.e. a stronger force is needed to stop a more massive object (please, correct me, if I'm not saying things right)


So, the big question: if an ant fell with the same velocity as a human being, would it still experience less damage on impact because it has less mass?
I know this question may not be that simple as also anatomy of an ant should be taken into account, but I'm trying to exemplify things to understand what happens. Thanks to whoever will help!
 A: Energy won't kill anyone. I could absorb 1 ton of TNT (4.184 GJ) worth of energy unharmed given enough time. The thing that hurts will be power or force. But experiencing massive g's won't harm you for short enough periods of time, so let's look at power. Assume both a human and ant have the same velocity immediately before impact. Then there is a linear relationship between ant energy to human energy, their mass ratio is the factor.
$$
\frac{E_a}{E_h} = \frac{m_a~v^2}{m_h~v^2} = \frac{m_a}{m_h}
$$
However, the problem comes in when the ant needs to stop vs the human needing to stop. If we assume the ant and human have similar enough geometry and density (they really don't but just so we don't base this off specific ant and person sizes) then their mass ratio is their volume ratio which is the cube of their length scaling ratio, seen below,
$$
\frac{m_a}{m_h} = \frac{\rho V_a}{\rho V_h} = \frac{{l_a}^3}{{l_h}^3} = \bigg(\frac{l_a}{l_h}\bigg)^3
$$
If you need to get rid of all that kinetic energy on impact, then you'll need to displace it in some length I've called $l$, but let's call the mass ratio $\mu=m_a/m_h$ and the length ratio will be $\mu^{1/3}$
If we want to displace $v$ in $l$ let's say at constant acceleration because this is the smallest maximum acceleration the body would experience for a drop, I understand this violates what I referenced, but let's assume $l$,$v$, and $a$ are such that the time is large enough to be dangerous for these accelerations.
$$
v^2 = 2al \\
a = \frac{v^2}{2l}
$$
Then we'll want the time from quadratic equation and we can compute average power
$$
-l = -vt+at^2 = vt + \frac{v^2}{2l}t^2, ~ t>0 \\
t = \frac{v \pm \sqrt{\strut v^2+4lv^2/2l}}{v^2/2l} \\
t = \frac{v \pm \sqrt{\strut 3v^2}}{v^2/2l} = \frac{2l}{v}(1 \pm \sqrt{3})\\
t = \frac{2l}{v}(1 + \sqrt{3})
$$
The power for the impact is the the energy dispersed over the time of impact: $P=E/t$
$$
P = \frac{mv^2}{2l(1+\sqrt{3})/v} = \frac{mv^3}{2l(1+\sqrt{3})}
$$
Comparing the ant to the human
$$
P_a = \frac{m_av^3}{2l_a(1+\sqrt{3})} \\
P_h = \frac{m_hv^3}{2l_h(1+\sqrt{3})} \\
\frac{P_a}{P_h} = \frac{m_a}{m_h}\frac{l_h}{l_a} = \mu~\mu^{-1/3} \\
\frac{P_a}{P_h} = \mu^{2/3}
$$
This just tells us that an ant uses less power to stop the fall, but if we compare to their individual masses, and call this the effective damage $D$
$$
\frac{P_a}{m_a} = D_a \\
\frac{P_h}{m_h} = D_h \\
\frac{D_a}{D_h} = \frac{P_a}{P_h}\frac{m_h}{m_a} = \mu^{2/3} \mu^{-1} = \mu^{-1/3} \\
D_a = D_h \mu^{-1/3}
$$
Assuming $m_a=10g$ and $m_h=70kg$, $\mu=143\cdot10^{-6}$ so $D_a/D_h \approx 19$. Telling us that an ant would have about 20 times the hurt. Using values found in Farcher's article, $m_a=0.3g$ and $m_h=82kg$, then $D_a/D_h \approx 65$.
But this is assuming that the time of impact will be large enough to be important, but this time is proportional to the length the body will use to land and inverse to the speed of impact, so if the ant travels quickly enough relative to this length, it will be too short of a time to be damaged. Since we're assuming both the ant and the human have the same speed, the human is less likely to fall into the short time range of impact durations.
