If a coin is thrown upwards from a height of 3.1 m, with an initial speed of 1 m/s, how many seconds does it remain in the air?

To determine how long the coin remains in the air, we need to find the time it takes to reach a height of 0 m (when it falls back down).

The height of the coin at any given time (t) can be determined by the equation:
h(t) = h_0 + v_0t - \frac{1}{2}gt^2
where:
- h(t) is the height of the coin at time t
- h_0 = 3.1 \, \text{m} is the initial height
- v_0 = 1 \, \text{m/s} is the initial speed
- g = 9.81 \, \text{m/s}^2 is the acceleration due to gravity

At the highest point, the vertical speed becomes 0, so:
v(t) = v_0 - gt = 0
Solving for t:
1 - 9.81t = 0
t = \frac{1}{9.81} \approx 0.102 \, \text{s}

The total time in the air is twice this time since it takes the same time to go up and to come down:
T_{\text{total}} = 2 \times 0.102 \, \text{s} = 0.204 \, \text{s}

\boxed{0.204 \, \text{s}} is the total time that the coin remains in the air.