一枚硬币被抛向空中 $t$ 秒后，其距离地面的高度为 $h$ 米，可以用 $h = -6t^2 + 24t + 12$ 表示。

1. Factor out the coefficient of $t^2$ from the $t$ terms

$h = -6(t^2 - 4t) + 12$

2. Complete the square for the expression inside the bracket

$h = -6\left[ (t - 2)^2 - (-2)^2 \right] + 12$
$h = -6\left[ (t - 2)^2 - 4 \right] + 12$

3. Expand and simplify to find the maximum height

$h = -6(t - 2)^2 + 24 + 12$
$h = -6(t - 2)^2 + 36$
Since $-6(t - 2)^2 \leqslant 0$ for all real $t$, the maximum value of $h$ is $36$.

1. Set the height to zero to find when the coin lands on the ground and simplify the equation.

When the coin lands on the ground, $h = 0$.
$-6t^2 + 24t + 12 = 0$
Divide by $-6$:
$t^2 - 4t - 2 = 0$

2. Solve the quadratic equation to find the exact values of $t$.

Using the quadratic formula:
$t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-2)}}{2(1)}$
$t = \frac{4 \pm \sqrt{16 + 8}}{2}$
$t = \frac{4 \pm \sqrt{24}}{2}$
$t = \frac{4 \pm 2\sqrt{6}}{2}$
$t = 2 \pm \sqrt{6}$

3. Reject the invalid root and state the final exact duration.

Since time $t$ must be positive ($t > 0$):
$t = 2 + \sqrt{6}$
(Reject $t = 2 - \sqrt{6}$ as it is negative)

The exact duration is $2 + \sqrt{6}$ seconds.