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Date	May 2022	Marks available	4	Reference code	22M.2.AHL.TZ2.6
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Determine	Question number	6	Adapted from	N/A

Question

At an archery tournament, a particular competition sees a ball launched into the air while an archer attempts to hit it with an arrow.

The path of the ball is modelled by the equation

$(\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 5 \\ 0 \end{matrix}) + t (\begin{matrix} u_{x} \\ u_{y} - 5 t \end{matrix})$

where $x$ is the horizontal displacement from the archer and $y$ is the vertical displacement from the ground, both measured in metres, and $t$ is the time, in seconds, since the ball was launched.

$u_{x}$ is the horizontal component of the initial velocity
$u_{y}$ is the vertical component of the initial velocity.

In this question both the ball and the arrow are modelled as single points. The ball is launched with an initial velocity such that $u_{x} = 8$ and $u_{y} = 10$ .

An archer releases an arrow from the point $(0, 2)$ . The arrow is modelled as travelling in a straight line, in the same plane as the ball, with speed $60 {m s}^{- 1}$ and an angle of elevation of $10 °$ .

Find the initial speed of the ball.

[2]

a.i.

Find the angle of elevation of the ball as it is launched.

[2]

a.ii.

Find the maximum height reached by the ball.

[3]

Assuming that the ground is horizontal and the ball is not hit by the arrow, find the $x$ coordinate of the point where the ball lands.

[3]

For the path of the ball, find an expression for $y$ in terms of $x$ .

[3]

Determine the two positions where the path of the arrow intersects the path of the ball.

[4]

Determine the time when the arrow should be released to hit the ball before the ball reaches its maximum height.

[4]

Markscheme

$\sqrt{10^{2} + 8^{2}}$ (M1)

$= 12.8 (12.8062 \dots, \sqrt{164}) (m s^{- 1})$ A1

[2 marks]

a.i.

$\tan^{- 1} (\frac{10}{8})$ (M1)

$= 0.896$ OR $51.3$ ( $0.896055 \dots$ OR $51.3401 \dots °$ ) A1

Note: Accept $0.897$ or $51.4$ from use of $arcsin (\frac{10}{12.8})$ .

[2 marks]

a.ii.

$y = t (10 - 5 t)$ (M1)

Note: The M1 might be implied by a correct graph or use of the correct equation.

METHOD 1 – graphical Method

sketch graph (M1)

Note: The M1 might be implied by correct graph or correct maximum (eg $t = 1$ ).

max occurs when $y = 5 m$ A1

METHOD 2 – calculus

differentiating and equating to zero (M1)

$\frac{d y}{d t} = 10 - 10 t = 0$

$t = 1$

$y (= 1 (10 - 5)) = 5 m$ A1

METHOD 3 – symmetry

line of symmetry is $t = 1$ (M1)

$y (= 1 (10 - 5)) = 5 m$ A1

[3 marks]

attempt to solve $t (10 - 5 t) = 0$ (M1)

$t = 2$ (or $t = 0$ ) (A1)

$x (= 5 + 8 \times 2) = 21 m$ A1

Note: Do not award the final A1 if $x = 5$ is also seen.

[3 marks]

METHOD 1

$t = \frac{x - 5}{8}$ M1A1

$y = (\frac{x - 5}{8}) (10 - 5 \times \frac{x - 5}{8})$ A1

METHOD 2

$y = k (x - 5) (x - 21)$ A1

when $x = 13, y = 5$ so $k = \frac{5}{(13 - 5) (13 - 21)} = - \frac{5}{64}$ M1A1

$(y = - \frac{5}{64} (x - 5) (x - 21))$

METHOD 3

if $y = a x^{2} + b x + c$

$0 = 25 a + 5 b + c$
$5 = 169 a + 13 b + c$
$0 = 441 a + 21 b + c$ M1A1

solving simultaneously, $a = - \frac{5}{64}, b = \frac{130}{64}, c = - \frac{525}{64}$ A1

( $y = - \frac{5}{64} x^{2} + \frac{130}{64} x - \frac{525}{64}$ )

METHOD 4

use quadratic regression on $(5, 0), (13, 5), (21, 0)$ M1A1

$y = - \frac{5}{64} x^{2} + \frac{130}{64} x - \frac{525}{64}$ A1

Note: Question asks for expression; condone omission of " $y =$ ".

[3 marks]

trajectory of arrow is $y = x \tan 10 + 2$ (A1)

intersecting $y = x \tan 10 + 2$ and their answer to (d) (M1)

$(8.66, 3.53) ((8.65705 \dots, 3.52647 \dots))$ A1

$(15.1, 4.66) ((15.0859 \dots, 4.66006 \dots))$ A1

[4 marks]

when $x_{target} = 8.65705 \dots, t_{target} = \frac{8.65705 \dots - 5}{8} = 0.457132 \dots s$ (A1)

attempt to find the distance from point of release to intersection (M1)

$\sqrt{8.65705 \dots^{2} + {(3.52647 \dots - 2)}^{2}} (= 8.79060 \dots m)$

time for arrow to get there is $\frac{8.79060 \dots}{60} = 0.146510 \dots s$ (A1)

so the arrow should be released when

$t = 0.311 (s) (0.310622 \dots (s))$ A1

[4 marks]

Examiners report

This question was found to be the most difficult on the paper. There were a good number of good solutions to parts (a) and part (b), frequently with answers just written down with no working. Part (c) caused some difficulties with confusing variables. The most significant difficulties started with part (d) and became greater to the end of the question. Few candidates were able to work through the final two parts.

a.i.

[N/A]

a.ii.

[N/A]

Syllabus sections

Topic 2—Functions » SL 2.4—Key features of graphs, intersections using technology

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Find the value of $n$ .
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16N.2.AHL.TZ0.H_5a:
Sketch the graph of $f$ indicating clearly any intercepts with the axes and the coordinates of any local maximum or minimum points.
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Sketch the graph of $y = f(x)$ showing clearly the equations of asymptotes and the coordinates of any intercepts with the axes.
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Find the largest possible domain $D$ for $f$ to be a function.
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Find the value of this minimum area.
19M.1.AHL.TZ1.H_8c:
find the value of $f\left( 4 \right)$ .
18M.2.AHL.TZ2.H_10b:
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17M.1.AHL.TZ1.H_11f:
Determine the area of the region enclosed between the graph of $y = f\left( {\left| x \right|} \right)$ , the $x$ -axis and the lines with equations $x = - 1$ and $x = 1$ .
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21M.1.SL.TZ1.11c:
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Find the values of $d$ where Scarlett cannot hear the siren.
21M.1.AHL.TZ2.5b:
Find the value of $k$ .
21N.2.AHL.TZ0.2a.ii:
minimum value of $h$ .
21N.1.AHL.TZ0.10b:
On the same set of axes draw the graph of $y = g (x)$ , showing any intercepts and asymptotes.
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Find the time, in seconds, it takes for the blade $[BC]$ to make one complete rotation under these conditions.
21N.2.SL.TZ0.3a.i:
maximum value of $h$ .
21N.2.SL.TZ0.3a.ii:
minimum value of $h$ .
21N.2.SL.TZ0.3b.i:
Find the time, in seconds, it takes for the blade $[BC]$ to make one complete rotation under these conditions.
21N.2.SL.TZ0.3b.ii:
Calculate the angle, in degrees, that the blade $[BC]$ turns through in one second.
21N.2.SL.TZ0.3c.ii:
Find the period of the function.
21N.2.SL.TZ0.3d:
Sketch the function $h (t)$ for $0 \leq t \leq 5$ , clearly labelling the coordinates of the maximum and minimum points.
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At any given instant, find the probability that point $C$ is visible from Tim’s window.
21N.2.AHL.TZ0.2a.i:
maximum value of $h$ .
21N.2.AHL.TZ0.2b.ii:
Calculate the angle, in degrees, that the blade $[BC]$ turns through in one second.
21N.2.AHL.TZ0.2c.i:
Write down the amplitude of the function.
21N.2.AHL.TZ0.2c.ii:
Find the period of the function.
21N.2.AHL.TZ0.2e.i:
Find the height of $C$ above the ground when $t = 2$ .
21N.2.AHL.TZ0.2e.ii:
Find the time, in seconds, that point $C$ is above a height of $100 m$ , during each complete rotation.
21N.2.AHL.TZ0.2f:
The wind speed increases and the blades rotate faster, but still at a constant rate.

Given that point $C$ is now higher than $110 m$ for $1$ second during each complete rotation, find the time for one complete rotation.

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