Disc golf plinko

The time taken for an object to fall, freely, a distance $h$ is $t = \sqrt{\dfrac{2h}{g}}$ .

Let the tree top be at height $D$ and the branch at height $x$ . Then $t_{top} = \sqrt{\dfrac{2D}{g}}$ .

To find $t_{max}$ , we note that the frisbee first falls a distance $D - x$ , hitting the branch and coming to a full stop, and then falls a further distance $x$ to the ground. So in this scenario, where the frisbee hits a single branch, the time $T(x)$ taken to reach the ground is

$T(x) = \sqrt{\dfrac{2(D - x)}{g}} + \sqrt{\dfrac{2x}{g}}$ .

We now need to find where $\dfrac{dT}{dx} = 0$ :

$\dfrac{dT}{dx} = (\dfrac{1}{\sqrt{2g}})(-\dfrac{1}{\sqrt{D - x}} + \dfrac{1}{\sqrt{x}}) = 0 \Longrightarrow D - x = x \Longrightarrow x = \dfrac{D}{2}$ .

Now $\dfrac{d^{2}T}{dx^{2}} \lt 0$ for $0 \lt x \lt D$ , so by the second derivative test we know that the critical point found above will result result in a maximum value for $T(x)$ , and thus

$t_{max} = T(\frac{D}{2}) = 2*\sqrt{\dfrac{D}{g}}$ .

Therefore, $\dfrac{t_{max}}{t_{top}} = \dfrac{2*\sqrt{\frac{D}{g}}}{\sqrt{\frac{2D}{g}}} = \sqrt{2} = \boxed{1.414}$

to $3$ decimal places.