What's this about kinetic energy increasing with the fifth power of length? I don't quite understand this quote from Stephen J. Gould's Ever since Darwin, where he talks about the compensating physical characteristics of organisms for their size. 

Other essential features of organisms change even more rapidly
  with increasing size than the ratio of surface to volume. Kinetic 
  energy, in some situations, increases as length raised to the ﬁfth
  power. If a child half your height falls down, its head will hit 
  with not half, but only 1/32 the energy of yours in a similar fall. In return,
  we are protected from the physical force of its tantrums, for 
  the child can strike with, not half, but only 1/32 of the energy 
  we can muster.

Mass increases with the third power of length, and even if we allow for a shorter fall (even though that is not clear from the wording) for the child (who is half as tall), I can only come up with the fourth power. 
 A: Model the child as a rod  of mass $m$ and length $l$ standing on the ground vertically, with center of mass at height $l/2$, with feet glued to the ground but the rest of the body able to rotate. 
When upright, the potential energy is $mgl/2$. When lying on the ground, the potential is 0. So when falling, the child hits the ground with energy $mgl/2$. Equating this with $\frac{1}{2}I\omega^2$, with $I=\frac{m l^2}{3}$ (the moment of inertia of a rigid rod rotating about its end) one gets 
$$\omega=\sqrt{\frac{3g}{l}}.$$
The head hits the ground with velocity $l\omega$. Letting the mass of the head be $m_h$, the kinetic energy of the head hitting the ground is 
$$T=\frac{1}{2}m_h(l\omega)^2=\frac{3}{2} g l m_h.$$
Since $l\rightarrow \lambda l$ and $m_h\rightarrow\lambda^3m_h$ under a scale transformation $\lambda$, we have 
$$T\rightarrow\lambda^4T.$$
So unless I'm overlooking something, I can't quite see where Gould's extra factor of $\lambda$ comes from. Assuming this is right and Gould is mistaken, I suspect he got the $\lambda^5$ figure by looking at $E=\frac{1}{2}I\omega^2$ and noting that $I\rightarrow\lambda^5I$, but forgot to look in detail as to how $\omega$ transforms.
NOTE: If it makes it easier to swallow, this analysis can also be carried out by assuming the child is a cylinder of radius $R$, length $L$ and constant density $\rho$ which is free to rotate about its base. The inertia tensor is 
$$\mathbf{I}=\left(
\begin{array}{ccc}
 \frac{1}{12} L \pi  R^2 \left(4 L^2+3 R^2\right) \rho  & 0 & 0 \\
 0 & \frac{1}{12} L \pi  R^2 \left(4 L^2+3 R^2\right) \rho  & 0 \\
 0 & 0 & \frac{1}{2} L \pi  R^4 \rho  \\
\end{array}
\right)$$
and repeating the previous analysis using $I=\mathbf{I}_{xx}$ and $m=\pi  L \rho  R^2$ gives a head energy which scales as
$$T=m_h(L\omega)^2=\frac{6 g L^3 m_h}{4 L^2+3 R^2}\rightarrow\lambda^4T$$
under scale transformation.
A: It appears that the statement is half-right....
As answered by @DumpsterDoofus, the factor for a scaled object falling a scaled height, is the fourth power.
However, if we scale an object that is traveling at a velocity of  $x \frac{\text{body-lengths}}{\text{second}}$, then a fifth power for the kinetic energy is correct. The size and thus mass in the kinetic energy equation scales as the cube, and the velocity term in the kinetic energy scales as the square.
The difference in the two scenarios comes from correctly not scaling the distance factor in the value of $g$.
