Let's suppose there is a straight rigid bar with height $h$ and center of mass at the middle of height $h/2$. Now if the bar is vertically upright from ground, how long will it take to fall on the ground and what is the equation of motion of the center of mass (Lagrangian)?
Answer
A falling tree is basically an inverted pendulum.
The period of a pendulum of length $h$ for small oscillations is $2\pi \sqrt{h/g}$, with $g$ the acceleration due to gravity, about $10\ m/s$. For an inverted pendulum near the top of its arc, there is no period, but the quantity $\sqrt{h/g}$ does represent a characteristic time scale for this system. The tree will take a few of these characteristic times to fall. $h$ for a tree is an "effective height", and depends on the mass distribution of the tree. If all the mass is at the top, $h$ is the height of the tree. If the tree is uniform, $h$ is $2/3$ the true height.
For a tree with $h = 40\ m$, the characteristic time is $2\ s$. For small angles, the angle the tree makes with the vertical will be multiplied by $e$ in this time. Let's start the tree at $1^{\circ}$ so that it needs to multiply its angle by $90$ to fall. $\ln(90) = 4.5$ so the tree takes about $9\ s$ to fall.
This is mathematically an underestimate because the characteristic time increases slightly as the tree falls, but not too much. Give it a nice round $10\ s$ and you get something that matches the first YouTube video I found.
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