A tank is firing rounds during a target practice. The height, $h$ metres, of a projectile above the ground during flight is governed by the equation $h = 2 + \frac{4}{5}t - \frac{1}{250}t^{2}$, where $t$ is the time in seconds for the projectile in flight.

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

$h = -\frac{1}{250}(t^{2} - 200t) + 2$

2. Complete the square for the expression inside the bracket.

$h = -\frac{1}{250}\left[ (t - 100)^{2} - 100^{2} \right] + 2$
$h = -\frac{1}{250}\left[ (t - 100)^{2} - 10000 \right] + 2$

3. Expand and simplify to obtain the final form.

$h = -\frac{1}{250}(t - 100)^{2} + \left(-\frac{1}{250}\right)(-10000) + 2$
$h = -\frac{1}{250}(t - 100)^{2} + 40 + 2$
$h = 42 - \frac{1}{250}(t - 100)^{2}$

1. Deduce the completed square form from part (a)

From part (a), $h = 42 - \frac{1}{250}(t - 100)^2$.

2. State the maximum height and the time it occurs

The maximum value of $h$ occurs when $(t - 100)^2 = 0$.
Maximum height = $42\text{ m}$
Time = $100\text{ s}$

1. Set up the inequality for the height being at least 32 metres

Using the completed square form from part (a), $h = 42 - \frac{1}{250}(t - 100)^2$.
We require $h \geqslant 32$:
$42 - \frac{1}{250}(t - 100)^2 \geqslant 32$

2. Solve the quadratic inequality for $t$

$10 \geqslant \frac{1}{250}(t - 100)^2$
$(t - 100)^2 \leqslant 2500$
$-50 \leqslant t - 100 \leqslant 50$
$50 \leqslant t \leqslant 150$

3. Calculate the length of time

Length of time $= 150 - 50 = 100$ seconds.