Very difficult calculus related rates question I was looking at a math problem from a few years ago that I could not solve. I was wondering if anyone knows where to even begin. I have the answer along with the question, however, I do not know how to arrive at this answer -
Here is the question:

The answer is:


To simplify  this problem, we can change the perspective by noting that climbing a mountain with decreasing velocity is equivalent to climb with constant velocity a mountain that grows larger as we rise up. In particular, based on the data of the problem, we can see our progressively enlarging mountain as a cylinder: in fact, since at any height  $z $ the corresponding radius of the cone is  $r_0\,(h-z)/z\,\, \,$, its circumference is $2\,\pi\,  r_0(h-z)/z \,\,\,$ and the velocity is $ v_0\,(h-z)/z\,\, \,$, the time needed by a climber to cover the circumference is $2\,\pi \, r_0/v_0\,\,\,$, i.e. is independent by the height $z $. In other words, we   can simplify this problem by imagining a man climbing a cilindrical mountain having radius $r_0$ with constant velocity  $v_0$. The problem therefore reduces to that of calculating the position of our original target point on such a cylinder. 
To do this, we can note that if we call $L$ the slant of the initial cone and $x$ the distance from its top at a given instant of our ascent, the instantaneous velocity is $v_0 \cdot \frac{x}{L} \,\,$. So, at any instant of our ascent, to cover an infinitesimal distance $dx$ we need a time equal to $\frac{L}{v_0\,x} dx \,$. Integrating in the range between $L$ and $L/2$ (i.e., from the beginning of the ascent to the height corresponding to our target point) and multiplying to $v_0$ to get the distance covered on the cylinder, this leads to a distance of 
$$\displaystyle \int_{L/2}^L \frac{L}{x} dx=L \left[\log L - \log (L/2) \right]=L \log 2$$  
Therefore, to reach the original target point climbing on the cylinder, unwrapping the lateral surface of the cylinder on to a plane to see it as a rectangle, we have to cover a distance equal to the hypothenuse of a right triangle whose legs are $L \log 2$ and $ \pi \, r_0$. This directly yields a distance of 
$$\sqrt {(L \cdot \log 2)^2 + (\pi r_0)^2}$$
which divided by the velocity $v_0$ (remind that climbing on our cylinder we have assumed the velocity to be constant) gives a time $T$ equal to
$$T= \sqrt {\frac {(L \cdot \log 2)^2}{v_0^2} + \frac{(\pi r_0)^2}{v_0^2}}$$
Because the slant $L$ of the initial cone is equal to $\sqrt{h^2 + r_0^2} \,\,\,$, we obtain
$$T=\sqrt {\frac {(h^2 + r_0^2) \cdot \log^2 2}{v_0^2} + \frac{(\pi r_0)^2}{v_0^2}}$$
and substituting $h=4 \,\,$, $r_0=100\,\,$, and $v_0=2\,\,$, we obtain for $T $ the value
$$\sqrt {\frac {(4^2 + 100^2) \cdot \log^2 2}{2^2} + \frac{(100 \pi)^2}{2^2}} \\ = \sqrt {2504 \cdot \log^2 2 + 2500 \pi^2}  $$
