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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

xy=50+tuxuy-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.

In this question both the ball and the arrow are modelled as single points. The ball is launched with an initial velocity such that ux=8 and uy=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 60m 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]
b.

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]
c.

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

[3]
d.

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

[4]
e.

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

[4]
f.

Markscheme

102+82           (M1)

=12.8   12.8062, 164 ms-1          A1

 

[2 marks]

a.i.

tan-1108           (M1)

=0.896   OR   51.3   (0.896055   OR   51.3401°)           A1


Note:
Accept 0.897 or 51.4 from use of arcsin1012.8.

 

[2 marks]

a.ii.

y=t10-5t           (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=5m           A1

METHOD 2 – calculus

differentiating and equating to zero           (M1)

dydt=10-10t=0

t=1

y=110-5=5m           A1

 

METHOD 3 – symmetry

line of symmetry is t=1           (M1)

y=110-5=5m           A1

 

[3 marks]

b.

attempt to solve t10-5t=0           (M1)

t=2  (or t=0)          (A1)

x =5+8×2= 21m           A1

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

 

[3 marks]

c.

METHOD 1

t=x-58            M1A1

y=x-5810-5×x-58           A1


METHOD 2

y=kx-5x-21           A1

when x=13, y=5 so k=513-513-21=-564            M1A1

y=-564x-5x-21

 

METHOD 3

if y=ax2+bx+c

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

solving simultaneously, a=-564, b=13064, c=-52564           A1

(y=-564x2+13064x-52564)

 

METHOD 4

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

y=-564x2+13064x-52564           A1


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

 

[3 marks]

d.

trajectory of arrow is y=xtan10+2             (A1)

intersecting y=xtan10+2 and their answer to (d)             (M1)

8.66, 3.53  8.65705, 3.52647           A1

15.1, 4.66     15.0859, 4.66006           A1

 

[4 marks]

e.

when xtarget=8.65705,  ttarget=8.65705-58=0.457132s             (A1)

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

8.657052+3.52647-22  =8.79060m

time for arrow to get there is 8.7906060=0.146510s             (A1)

so the arrow should be released when

t=0.311s  0.310622s           A1 

 

[4 marks]

f.

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]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.
[N/A]
f.

Syllabus sections

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