Find the initial speed of the ball.

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

Find the maximum height reached by the ball.

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.

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

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

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

$\sqrt{{10}^2+8^2}$           (M1)

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

[2 marks]

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

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

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

[2 marks]

$y=t\left(10-5t\right)$           (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 $y=5 \text{m}$           A1

METHOD 2 – calculus

differentiating and equating to zero           (M1)

$\frac{dy}{dt}=10-10t=0$

$t=1$

$y\left(=1\left(10-5\right)\right)=5 \text{m}$           A1

METHOD 3 – symmetry

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

$y\left(=1\left(10-5\right)\right)=5 \text{m}$           A1

[3 marks]

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

$t=2$  (or $t=0$)          (A1)

$x \left(=5+8\times 2\right)= 21 \text{m}$           A1

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

[3 marks]

METHOD 1

$t=\frac{x-5}{8}$            M1A1

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

METHOD 2

$y=k\left(x-5\right)\left(x-21\right)$           A1

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

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

METHOD 3

if $y=a{x}^2+bx+c$

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

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

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

[4 marks]

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

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

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

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

so the arrow should be released when

$t=0.311 \left(\text{s}\right) \left(0.310622\dots \left(\text{s}\right)\right)$           A1 

[4 marks]

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.