Mathematics · Oct / Nov 2022 · Paper 51 · Question 5
YMA01
As printed in the paper
The question as text
5. A small ball is projected vertically upwards with speed 29.4 m s–1 from a point A which is 19.6 m above horizontal ground. The ball is modelled as a particle moving freely under gravity until it hits the ground. It is assumed that the ball does not rebound.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
5(a)[3]
0
=
14.7
2
−
2gs
22 or 22.1 (m)
1
19.6
=
29.4
+
2
t
gt
2
1
N.B.
19.6
=
29.4
−
2
is M0A0
t
gt
2
5(b)[3]
1
−
19.6
=
29.4
+
2
is M0A0
t
gt
2
1
2
is M0A0 unless they go on to subtract 6
−
19.6
=
29.4
t
−
gt
2
from the positive root
t = 0.61 or 0.606 (s)
v
5(c)[3]
29.4
0 t
3
Notes for question 5
M1 Complete method to find distance UP
N.B. They may find time UP (1.5s ) AND use it to find distance UP
5(a)
OR: (Distance from A to top – Distance from ‘14.7’ to top)
= (44.1 – 33.075)
A1 Correct equation(s) used
A1 cao
M1 Complete method to find required time
N.B. They may find the speed as it hits the ground (
13
=
35.334...
)
5(b)
g
AND use it to find the time.
A1 Correct equation(s) used
A1 cao
N.B. If they add to or subtract from 0.606, it’s M0 for an incorrect
method.
B1 V shape with v coord of end pt > 29.4 and each half roughly equally
5(c)
inclined to the taxis. B0 if a vertical line is included at the end.
B1 29.4 independent
B1 3 independent
( 3
−
+
2 )
j +(
+
j = ( 3
)
−+
)
+
(2
+
)
j
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