Mathematics · Oct / Nov 2025 · Paper 51 · Question 4
YMA01
As printed in the paper
The question as text
4. The point A is 10 m above horizontal ground. At time t = 0, a particle P is projected vertically upwards with speed 5 m s –1 from A. Particle P moves freely under gravity.
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4(a)[3]
Correct use of suvat to give an
equation in h only.
=
−
e.g.
0
5
2
2gh
=
−
or
25
0
2
gh
( )
1.3 or 1.28 m
4(b)[4]
Use suvat to form an expression
for the distance travelled by
either particle in terms of T or t.
1
e.g.
±
−
2
Case 1 (P)
(5
T
gT
)
2
1
(
)
(
)
±
−
−
−
2
7
T
1
g T
1
Case 1 (Q)
2
OR
1
±
5(
t
+
1)
−
g t
(
+
1)
2
Case 2 (P)
2
1
Case 2 (Q)
±
(7
t
−
gt
2
)
2
Use suvat to form a correct
expression, in terms of the same
variable, for the distance
travelled by the other particle
1
1
(
)
(
)
10
+
5
T
−
gT
2
=
7
T
−
1
−
g T
−
1
2
2
1
1
Or
10
+
5(
t
+
1)
−
g t
(
+
1)
2
=
7
t
−
gt
2
2
T =
(
) 1.9 or 1.86
ALT 1
Find position and velocity of P at t
= 1:
10.1 m above ground, moving
downwards at 4.8
m s− then:
1
1
(P) M1
±
+
2
(4.8
t
gt
)
2
1
(Q) M1
±
(7
t
−
gt
2
)
2
1
1
+
2
+
−
2
=
A1
(4.8
t
gt
)
(7
t
gt
)
10.1
2
2
A1
(
T =
) 1.9 or 1.86
ALT 2
Find position and velocity of Q at
t = 0:
11.9 m ‘below’ ground, moving
upwards at 16.8
m s− then:
1
1
±
−
2
(P) M1
(5
T
gT
)
2
1
(Q) M1
±
(16.8
T
−
gT
2
)
2
1
1
−
−
−
=
A1
16.8
T
gT
2
(5
t
gT
2
)
21.9
2
2
(
T =
) 1.9 or 1.86
4(c)[2]
e.g.
or
v
= −+
5
gT
v
=
5
−
gT
(
)
Speed of P =
ms−
13 or 13.2
1
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