Solved: At an archery tournament, a particular competition sees a ball launched into the... [IB Mathematics Applications & Interpretation (Math AI)]

Question

HLPaper 2

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 $\binom{x}{y} = \binom{5}{0} + t \binom{u_x}{u_y - 5t}$ 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 $u_y = 10$

Find the initial speed of the ball.

[2]

Verified

Solution

$\sqrt{10^2+8^2}$

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

[2]

Verified

Solution

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

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

Find the maximum height reached by the ball.

[3]

Verified

Solution

$y=t(10-5t)$

Note: The $\boldsymbol{M1}$ might be implied by a correct graph or use of the correct equation. METHOD 1 - graphical Method sketch graph Note: The $\boldsymbol{M1}$ might be implied by correct graph or correct maximum (eg $t=1$ ). max occurs when $y=5$ m

METHOD 2 - calculus

differentiating and equating to zero $\frac{\mathrm{d}y}{\mathrm{d}t}=10-10t=0$ $t=1$ $y(=1(10-5))=5$ m

METHOD 3 - symmetry

line of symmetry is $t=1$ $y(=1(10-5))=5$ m

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]

Verified

Solution

attempt to solve $t(10-5t)=0$ $t=2 \quad$ (or $t=0$ ) (A1) $x(=5+8\times 2)=21$ m Note: Do not award the final $\boldsymbol{A1}$ if $x=5$ is also seen.

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

[3]

Verified

Solution

METHOD 1 $t=\frac{x-5}{8}$ M1A1 $y=\left(\frac{x-5}{8}\right)\left(10-5\times \frac{x-5}{8}\right)$

METHOD 2

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

METHOD 3

if $y=ax^2+bx+c$ $0=25a+5b+c$ $5=169a+13b+c$ $0=441a+21b+c$ solving simultaneously, $a=-\frac{5}{64}, b=\frac{130}{64}, c=-\frac{525}{64}$ $\left(y=-\frac{5}{64}x^2+\frac{130}{64}x-\frac{525}{64}\right)$

METHOD 4

use quadratic regression on $(5,0),(13,5),(21,0)$ $y=-\frac{5}{64}x^2+\frac{130}{64}x-\frac{525}{64}$ A1

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

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

[4]

Verified

Solution

trajectory of arrow is $y=x\tan 10+2$ (A1) intersecting $y=x\tan 10+2$ and their answer to (d) (M1) $(8.66,3.53)((8.65705 \ldots, 3.52647 \ldots))$ A1 $(15.1,4.66) \quad ((15.0859 \ldots, 4.66006 \ldots))$ A1

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

[4]

Verified

Solution

when $x_{\text{target}}=8.65705 \ldots, t_{\text{target}}=\frac{8.65705 \ldots-5}{8}=0.457132 \ldots$ s attempt to find the distance from point of release to intersection $\sqrt{8.65705 \ldots^2+(3.52647 \ldots-2)^2}(=8.79060 \ldots$ m) time for arrow to get there is $\frac{8.79060 \ldots}{60}=0.146510 \ldots$ s so the arrow should be released when $t=0.311$ (s) $(0.310622 \ldots(\text{s}))$

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