一石头被竖直向上抛出，在被抛出 $t$ 秒后，它距离地面的高度 $h$（米）由公式 $h = -9t^2 + 12t + 1$ 给出。

1. Identify the physical meaning of the constant term by considering $t=0$

When $t = 0$, $h = 1$. The constant term represents the height of the stone at the start.

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

$h = -9\left(t^2 - \frac{12}{9}t\right) + 1$
$h = -9\left(t^2 - \frac{4}{3}t\right) + 1$

2. Complete the square inside the bracket

$h = -9\left[ \left(t - \frac{2}{3}\right)^2 - \left(-\frac{2}{3}\right)^2 \right] + 1$
$h = -9\left[ \left(t - \frac{2}{3}\right)^2 - \frac{4}{9} \right] + 1$

3. Expand and simplify to reach the final form

$h = -9\left(t - \frac{2}{3}\right)^2 + (-9)\left(-\frac{4}{9}\right) + 1$
$h = -9\left(t - \frac{2}{3}\right)^2 + 4 + 1$
$h = -9\left(t - \frac{2}{3}\right)^2 + 5$

1. Obtain the completed square form of $h$ from part (b)

$h = -9\left(t^2 - \frac{4}{3}t\right) + 1$
$h = -9\left(t - \frac{2}{3}\right)^2 + 4 + 1$
$h = -9\left(t - \frac{2}{3}\right)^2 + 5$

2. Deduce the maximum height and the time it occurs from the vertex form

Since $-9\left(t - \frac{2}{3}\right)^2 \leqslant 0$ for all real $t$, the maximum value of $h$ is $5$.
This occurs when $t - \frac{2}{3} = 0 \implies t = \frac{2}{3}$.