An object was launched from a machine. Its height, $y$ metres above ground, $t$ seconds after launch, can be modelled by a quadratic function $y = f(t)$. Four seconds after launch, the object reached its maximum height of 81 m and then dropped to a height of 54 m after a further three seconds.

1. Formulate the quadratic function using the given maximum point

Since the maximum height is $81\text{ m}$ at $t = 4$, the vertex of the parabola is $(4, 81)$.
Let the quadratic function be $y = a(t - 4)^2 + 81$.

2. Use the second given point to find the constant $a$

The object drops to $54\text{ m}$ after a further $3$ seconds, so at $t = 4 + 3 = 7$, $y = 54$.
Substitute $(7, 54)$ into the equation:
$54 = a(7 - 4)^2 + 81$
$54 = 9a + 81$
$9a = -27$
$a = -3$
So, the function is $y = -3(t - 4)^2 + 81$.

3. Calculate the initial height by substituting $t = 0$

The initial height is when $t = 0$:
$y = -3(0 - 4)^2 + 81$
$y = -3(16) + 81$
$y = -48 + 81 = 33$

1. Explain the use of symmetry to find the other time the object is at 54 m

The graph of a quadratic function is a parabola, which is symmetrical about its maximum point at $t = 4$.
Since the object is at a height of 54 m at $t = 4 + 3 = 7$ seconds, by symmetry, it must also be at 54 m at $t = 4 - 3 = 1$ second.

2. Calculate the duration the object stayed above 54 m

Time spent above 54 m $= 7 - 1 = 6$ seconds.