Solution 1:

After losing the kettle, people continue to go upstream for 20 minutes, and the kettle goes downstream: speed sum = people's countercurrent speed+water flow speed = people's still water speed-water flow speed+water flow speed = people's still water speed, and the distance between people and the kettle = their speed and time =20× people's still water speed. When the man found that the kettle was lost, he went back and swam down with it. The speed difference is between them. Catch-up time =2÷ water speed, so there are: 20× human still water speed =2÷ water speed × human still water speed, so water speed =110, catch-up time =2÷ water speed =20 minutes.

Solution 2:

Man speed: x (m/min) water speed: y (m/min) pan chasing time: z (min)

Distance for a person to swim for 20 minutes: 20(x-y)

The distance that the pot floats after people leave: 20 years.

Distance from the beginning of pot chasing to the end of pot chasing: z(x+y)

The floating distance of the pot after people start chasing it: zy

Then the following relationship holds: z(x+y)=20(x-y )+20y +zy.

After simplification: zx=20x

Z=20 (minutes)

So it took him 20 minutes to come back and look for the jar.

Solution 3:

The problems of meeting and chasing in flowing water and meeting and chasing and compounding can be solved in still water without water speed. Regardless of the water speed, the condition of "catching up in 2 kilometers" is useless. He swam forward for 20 minutes, so it only takes 20 minutes to come back.

Question 2 is obviously lacking in conditions. ......