Net force

The answers you quoted does NOT give a complete explanation to the question, and the writer showed incorrect concepts in the remarks.

The remarks should be rewritten as :
Fnet = 0 N ⇒ constant velocity or at rest continuously (a right concept)
constant velocity or at rest continuously ⇒ Fnet = 0 N (a right concept)
at rest momentarily ⇒ Fnet = 0 N (a wrong concept)

Now, John is "momentarily" at rest at C, and there is a non-zero net force acting on him.

Note the movement of John :
1. At the moment before C, John is falling downwards.
2. At C, John is "momentarily" at rest.
3. At the moment after C, John is bounced upwards.

Referring to 1, 2 and 3, you can see that John is accelerating upwards. This means there is an upward net force acting on John. The upward net force is equal to the difference between the tension in the elastic cord and the weight of the body.

2015-08-01 11:48:35 補充：
Think about this example:

A stone is projected vertically upwards, and gravitational force is acting on it. Therefore, the stone accelerates downwards, such that it firstly goes up, then is "momentarily at rest" at the highest point, and then falls down.

The crucial point lies on the word "momentarily" at rest. You should differentiate between "momentarily at rest" and "permanently at rest". The two situations are completely different.

It is just simple logic, an object is permanently at rest indicates that there is no (net) force acting on it. An object is momentarily at rest only stays at rest at that instant of time. It doesn't not indicates that there is no (net) force. Should there be no (net) force, the object would not move at all afterwards, and would become permanently at rest. This contradicts what is given in the question that John will bounce up afterwards.

A car just starts up to move is momentarily at rest, but there is a foce, produced by the engine, to pull the car forward.