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

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An archer shoots an arrow. The height, H metres, of the arrow above the ground is modelled by the formula $$H = 1.8 + 0.4d - 0.002d^{2}, \, d > 0$$ where d is the... 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, set the height H to 0:

0=1.8+0.4d0.002d20 = 1.8 + 0.4d - 0.002d^{2}

Rearranging gives:

0.002d20.4d1.8=00.002d^{2} - 0.4d - 1.8 = 0

Using the quadratic formula, where a = 0.002, b = -0.4, and c = -1.8:

d=b±b24ac2a d = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}

Calculating the discriminant:

b24ac=(0.4)24(0.002)(1.8)=0.16+0.0144=0.1744 b^{2} - 4ac = (-0.4)^{2} - 4(0.002)(-1.8) = 0.16 + 0.0144 = 0.1744

Now solving for d:

d=0.4±0.17440.004=0.4±0.41770.004 d = \frac{0.4 \pm \sqrt{0.1744}}{0.004} = \frac{0.4 \pm 0.4177}{0.004}

This provides two solutions, but we only consider the positive:

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 in the formula 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 starts at a height of 1.8 metres off the ground.

Step 3

Write 1.8 + 0.4d - 0.002d^{2} in the form A - B(d - C)^{2}

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Answer

To rewrite the equation in the desired form, we first complete the square:

Starting with:

H=1.8+0.4d0.002d2H = 1.8 + 0.4d - 0.002d^{2}

Factor out -0.002:

H=0.002(d2200d)+1.8H = -0.002(d^{2} - 200d) + 1.8

Next, complete the square for d2200dd^{2} - 200d:

=0.002((d100)210000)+1.8= -0.002((d - 100)^{2} - 10000) + 1.8

Distributing -0.002:

H=0.002(d100)2+(1.8+20)=0.002(d100)2+21.8H = -0.002(d - 100)^{2} + (1.8 + 20) = -0.002(d - 100)^{2} + 21.8

Thus, we have:

A=21.8,B=0.002,C=100A = 21.8, B = 0.002, C = 100.

Step 4

the maximum height of the arrow above the ground.

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Answer

For the adapted model, the maximum height can be found by evaluating at the vertex of the parabola described by the formula:

H=2.1+0.4d0.002d2H = 2.1 + 0.4d - 0.002d^{2}

The vertex occurs at:

d=b2a=0.42(0.002)=100extm d = -\frac{b}{2a} = -\frac{0.4}{2(-0.002)} = 100 ext{ m}

Evaluating H at d = 100:

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

Substituting:

H=2.1+40200=42.1extmH = 2.1 + 40 - 200 = 42.1 ext{ m}

Step 5

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

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Answer

The horizontal distance when the arrow reaches its maximum height is already calculated as:

d=100extmd = 100 ext{ m}

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