[Maximum mark: 9]
A mortar shell is fired from the top of a tower. The height, $h$, in metres of the mortar shell above the ground is modelled by the function
\[
h(t)=-2 t^{2}+20 t+8, \quad t \geq 0
\]
where $t$ is the time, in seconds, since the moment the mortar shell was fired.
(a) Write down the height of the tower.
[1]
(b) Determine the height of the mortar shell 5 seconds after it was fired.
[2]
The mortar shell hits its target on the ground $n$ seconds after it was fired.
(c) Find the value of $n$.
[2]
(d) Using a suitable sketch graph find the maximum height reached by the mortar shell and the time this is reached.
[2]
(e) Find an un-simplified expression for $h(t+2)$.
[1]
(f) Describe the transformation from $h(t)$ to $h(t+2)$.
[1]
Answer
Transformation: \text{Horizontal shift 2 units to the left}
Step 1 ：\(h(t) = -2t^2 + 20t + 8\)
Step 2 ：\(h(0) = -2(0)^2 + 20(0) + 8 = \boxed{8}\)
Step 3 ：\(h(5) = -2(5)^2 + 20(5) + 8 = \boxed{58}\)
Step 4 ：\(h(t) = 0 \Rightarrow -2t^2 + 20t + 8 = 0\)
Step 5 ：\(n = \{5 - \sqrt{29}, 5 + \sqrt{29}\}\)
Step 6 ：\(t_{vertex} = \frac{-b}{2a} = \frac{-20}{2(-2)} = 5\)
Step 7 ：\(h_{max} = h(5) = \boxed{58}\)
Step 8 ：\(h(t+2) = -2(t+2)^2 + 20(t+2) + 8\)
Step 9 ：\(h(t+2) = 20t - 2(t+2)^2 + 48\)
Step 10 ：Transformation: \text{Horizontal shift 2 units to the left}
