Differentiate between progressive waves and stationary waves, and solve a physics problem involving a falling child.

The child falls with an initial velocity of $u_c = 0 \text{ m/s}$.
The distance fallen by the child after time $t$ is $d_c = u_c t + \frac{1}{2}gt^2$.

$$d_c = 0 \cdot t + \frac{1}{2}(9.8)t^2$$ $$d_c = 4.9t^2$$

The height of the child from the ground at time $t$ is $h_c = H - d_c$.

$$h_c = 98 - 4.9t^2$$

The 'Supper man' starts flying 2 seconds after the child falls.
So, the time the 'Supper man' flies is $t_{SM} = t - 2$ (for $t \ge 2$).
The 'Supper man' flies at a constant speed of $v_{SM} = 39.2 \text{ m/s}$.
The distance covered by the 'Supper man' is $d_{SM} = v_{SM} \cdot t_{SM}$.

$$d_{SM} = 39.2(t-2)$$

The height of the 'Supper man' from the ground at time $t$ is $h_{SM} = H - d_{SM}$.

$$h_{SM} = 98 - 39.2(t-2)$$

The 'Supper man' catches the child when their heights are equal, i.e., $h_c = h_{SM}$.

$$98 - 4.9t^2 = 98 - 39.2(t-2)$$ $$4.9t^2 = 39.2(t-2)$$ $$4.9t^2 = 39.2t - 78.4$$

Rearrange into a quadratic equation:

$$4.9t^2 - 39.2t + 78.4 = 0$$

Divide the entire equation by 4.9:

$$\frac{4.9t^2}{4.9} - \frac{39.2t}{4.9} + \frac{78.4}{4.9} = 0$$ $$t^2 - 8t + 16 = 0$$

This is a perfect square trinomial:

$$(t-4)^2 = 0$$ $$t = 4 \text{ s}$$

The 'Supper man' catches the child after $\boxed{4 \text{ seconds}}$ from when the child started falling.

Substitute $t=4 \text{ s}$ into the child's height equation:

$$h_c = 98 - 4.9t^2$$ $$h_c = 98 - 4.9(4)^2$$ $$h_c = 98 - 4.9(16)$$ $$h_c = 98 - 78.4$$ $$h_c = 19.6 \text{ m}$$

The child is caught at a height of $\boxed{19.6 \text{ m}}$ from the ground.