An archer shoots an arrow. The height, $H$ metres, of the arrow above the ground is modelled by the formula $$H = 1.8 + 0.4d - 0.002d^2,$$ where $d > 0$ is the horizontal distance of the arrow from the archer, measured in metres. 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 $A - Bd - C d^2$ where $A, B$ and $C$ are constants to be found. (d) It is decided that the model should be adapted for a different archer. The adapted formula for this archer is $$H = 2.1 + 0.4d - 0.002d^2,$$ where $d o 0.$ 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.

A-Level Edexcel Maths Pure Parametric Equations: An archer shoots an arrow. The

An archer shoots an arrow - Edexcel - A-Level Maths Pure - Question 13 - 2017 - Paper 1

Question 13

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

Worked Solution & Example Answer:An archer shoots an arrow - Edexcel - A-Level Maths Pure - Question 13 - 2017 - Paper 1

Step 1

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

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Answer

To determine the horizontal distance when the arrow hits the ground, we set the height $H$ to zero in the formula:

$0 = 1.8 + 0.4d - 0.002d^2.$
Rearranging gives:

$0.002d^2 - 0.4d - 1.8 = 0.$
Using the quadratic formula, $d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 0.002$ , $b = -0.4$ , and $c = -1.8$ :

$d = \frac{0.4 \pm \sqrt{(-0.4)^2 - 4(0.002)(-1.8)}}{2(0.002)}.$
Solving the expression yields:

$d \approx 204 \text{ metres}.$

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 at the moment it is shot. This indicates that the arrow is launched from a height of 1.8 meters.

Step 3

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

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Answer

To rewrite the expression in the desired form:

$1.8 + 0.4d - 0.002d^2 = 1.8 - (-0.4)d - (-0.002)d^2.$
Thus, we identify $A = 1.8$ , $B = -0.4$ , and $C = -0.002$ .

Step 4

the maximum height of the arrow above the ground.

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Answer

For the new model $H = 2.1 + 0.4d - 0.002d^2$ , we need to find the maximum height. This occurs at the vertex of the parabola, given by:

$d = -\frac{b}{2a} = -\frac{0.4}{2(-0.002)} = 100 ext{ metres}.$
Substituting $d = 100$ back into the height equation:

$H = 2.1 + 0.4(100) - 0.002(100)^2 = 2.1 + 40 - 20 = 22.1 ext{ metres}.$

Step 5

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

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Answer

As determined previously, the horizontal distance at maximum height is $d = 100$ metres.

An archer shoots an arrow - Edexcel - A-Level Maths Pure - Question 13 - 2017 - Paper 1

Worked Solution & Example Answer:An archer shoots an arrow - Edexcel - A-Level Maths Pure - Question 13 - 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 - Bd - C d^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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