10 1
7.7735026919
* If he performs the operation zero times, he will reach the ground at time $1\times 0+\frac{10}{\sqrt{1}} = 10$.
* If he performs the operation once, he will reach the ground at time $1\times 1+\frac{10}{\sqrt{2}} \fallingdotseq 8.07$.
* If he performs the operation twice, he will reach the ground at time $1\times 2+\frac{10}{\sqrt{3}} \fallingdotseq 7.77$.
* If he performs the operation three times, he will reach the ground at time $1\times 3+\frac{10}{\sqrt{4}} = 8$.
Performing the operation four or more times will only delay the time to reach the ground. Therefore, it is optimal to perform the operation twice before jumping off, and the answer is $2+\frac{10}{\sqrt{3}}$.5 10
5.0000000000 It is optimal not to perform the operation at all.
1000000000000000000 100
8772053214538.5976562500
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