[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
$h (t) = - 2 t^{2} + 20 t + 8, t \geq 0$
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

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Answer

Transformation: \text{Horizontal shift 2 units to the left}

Steps

Step 1 ： $h (t) = - 2 t^{2} + 20 t + 8$

Step 2 ： $h (0) = - 2 (0)^{2} + 20 (0) + 8 = 8$

Step 3 ： $h (5) = - 2 (5)^{2} + 20 (5) + 8 = 58$

Step 4 ： $h (t) = 0 \Rightarrow - 2 t^{2} + 20 t + 8 = 0$

Step 5 ： $n = {5 - \sqrt{29}, 5 + \sqrt{29}}$

Step 6 ： $t_{v e r t e x} = \frac{- b}{2 a} = \frac{- 20}{2 (- 2)} = 5$

Step 7 ： $h_{m a x} = h (5) = 58$

Step 8 ： $h (t + 2) = - 2 (t + 2)^{2} + 20 (t + 2) + 8$

Step 9 ： $h (t + 2) = 20 t - 2 (t + 2)^{2} + 48$

Step 10 ：Transformation: \text{Horizontal shift 2 units to the left}