
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Let the height of the hole be $H$. Then the speed of Winnie-the-Pooh before hitting the first time is $V_{I}=\sqrt{2 g H}$.

In the second case: before hitting, Pooh has a speed $V_{1}=\sqrt{2 g h}$, where $h = H / n$. After the hit, the speed becomes $V_{1}^{\prime}=\frac{1}{k} \sqrt{2 g h}$. Approaching the ground, the speed is $V_{II}=\sqrt{\frac{1}{k^{2}} 2 g h + 2 g(H-h)}$.

The damage received by Pooh the first time is $I_{I}=M V_{I} \tau$, where $M$ is the mass of the bear and $\tau$ is the duration of the hit. The second time, the damage is $I_{II}=M \tau\left[\left(V_{1}-V_{1}^{\prime}\right)+V_{II}\right]$.

Finally, $I_{II} / I_{I}=\frac{k-1}{\sqrt{n} k} + \sqrt{\frac{(n-1) k^{2} + 1}{n k^{2}}} = \frac{5}{4}$.

Present the answer in LaTex format: \boxed{Your answer}
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