An archer shoots an arrow. The height, H metres, of the arrow above the ground is modelled by the formula $$H = 1.8 + 0.4d - 0.002d^2,$$ d > 0 Given that the arrow travels in a vertical plane until it hits the ground, (a) find the horizontal distance travelled by the arrow, as given by this model. (b) With reference to the model, interpret the significance of the constant 1.8 in the formula. (c) Write $$1.8 + 0.4d - 0.002d^2$$ in the form $$A - B(d - C)^2$$ where A, B and C are constants to be found. It is decided that the model should be adapted for a different archer. The adapted formula for this archer is $$H = 2.1 + 0.4d - 0.002d^2,$$ d > 0 Hence or otherwise, find, for the adapted model d(i) the maximum height of the arrow above the ground. (ii) the horizontal distance, from the archer, of the arrow when it is at its maximum height.

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An archer shoots an arrow - Edexcel - A-Level Maths Pure - Question 12 - 2017 - Paper 1

Question 12

An archer shoots an arrow. The height, H metres, of the arrow above the ground is modelled by the formula $$H = 1.8 + 0.4d - 0.002d^2,$$ d > 0 Given that the arr... show full transcript

Worked Solution & Example Answer:An archer shoots an arrow - Edexcel - A-Level Maths Pure - Question 12 - 2017 - Paper 1

Step 1

find the horizontal distance travelled by the arrow, as given by this model.

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Answer

To find the horizontal distance where the arrow hits the ground, we set the height $H$ to 0:

$0 = 1.8 + 0.4d - 0.002d^2$

Rearranging gives:

$0.002d^2 - 0.4d - 1.8 = 0$

Using the quadratic formula, we get:

$d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where:

$a = 0.002$
$b = -0.4$
$c = -1.8$

Calculating:

$d = \frac{-(-0.4) \pm \sqrt{(-0.4)^2 - 4 \cdot 0.002 \cdot (-1.8)}}{2 \cdot 0.002}$

$d = \frac{0.4 \pm \sqrt{0.16 + 0.0144}}{0.004}$

$d = \frac{0.4 \pm \sqrt{0.1744}}{0.004}$

$d \approx \frac{0.4 \pm 0.4176}{0.004}$

Taking the positive root:

$d \approx \frac{0.4 + 0.4176}{0.004} \approx 204$

Thus, the horizontal distance is approximately 204 metres.

Step 2

With reference to the model, interpret the significance of the constant 1.8 in the formula.

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Answer

The constant 1.8 represents the initial height of the arrow above the ground when the horizontal distance $d$ is zero. This means that when the arrow is shot, it begins at a height of 1.8 metres.

Step 3

Write 1.8 + 0.4d - 0.002d^2 in the form A - B(d - C)^2 where A, B and C are constants to be found.

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Answer

To rewrite the equation, we complete the square for the quadratic expression:

Starting with:

$H = 1.8 + 0.4d - 0.002d^2$

Factoring out -0.002 gives us:

$H = -0.002(d^2 - 200d) + 1.8$

Now, completing the square:

$H = -0.002((d - 100)^2 - 10000) + 1.8$

Expanding yields:

$H = -0.002(d - 100)^2 + 22.1$

Thus, we have:

$A = 22.1$
$B = 0.002$
$C = 100$ .

Step 4

the maximum height of the arrow above the ground.

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Answer

To find the maximum height for the adapted model:

Using the adapted formula, we have:

$H = 2.1 + 0.4d - 0.002d^2$

Using the previous steps, the vertex form gives the maximum height as:

At $d = 100$ :

$H = 2.1 + 0.4(100) - 0.002(100^2)$

Calculating:

$H = 2.1 + 40 - 20 = 22.1$

Therefore, the maximum height is 22.1 metres.

Step 5

the horizontal distance, from the archer, of the arrow when it is at its maximum height.

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Answer

From the calculations above, at maximum height, the horizontal distance, $d$ , is 100 metres, as we set it to reach the vertex of the quadratic function.

An archer shoots an arrow - Edexcel - A-Level Maths Pure - Question 12 - 2017 - Paper 1

Worked Solution & Example Answer:An archer shoots an arrow - Edexcel - A-Level Maths Pure - Question 12 - 2017 - Paper 1

find the horizontal distance travelled by the arrow, as given by this model.

With reference to the model, interpret the significance of the constant 1.8 in the formula.

Write 1.8 + 0.4d - 0.002d^2 in the form A - B(d - C)^2 where A, B and C are constants to be found.

the maximum height of the arrow above the ground.

the horizontal distance, from the archer, of the arrow when it is at its maximum height.

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