Edexcel International A Level Maths · M2 · Mechanics 2
Projectiles
28 free, high-quality past-paper-style questions on Projectiles for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 260 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-M2-1-B-01M2 · A2Easy6 marks
A marble rolls off the edge of a horizontal table top with speed 2.5 m s−1. The table top is 0.98 m above a horizontal floor. The marble is modelled as a particle moving freely under gravity from the moment it leaves the table. Take g = 9.8 m s−2.
Figure: the marble leaves the edge of the table horizontally; not to scale.
(a)
Find the time taken for the marble to reach the floor.
(2)
(b)
Find the horizontal distance from the edge of the table to the point where the marble hits the floor.
(1)
(c)
Find the speed of the marble as it hits the floor.
A pebble is flicked from a point O on horizontal ground with speed 12 m s−1 at an angle α above the horizontal, where tan α = 4/3. The pebble is modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.
Figure: the pebble projected from O at angle α to the horizontal, where tan α = 4/3; not to scale.
(a)
Find the time taken for the pebble to reach its greatest height.
(2)
(b)
Find the greatest height of the pebble above the ground.
(2)
(c)
Find the distance from O to the point where the pebble first hits the ground.
(2)
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Answer
Mark
Notes
Part (a)
vertical component 12 × 0.8 = 9.6 m s−1; at the top 0 = 9.6 − 9.8t
M1
t = 0.980 s (accept 0.98 s)
A1
Part (b)
0 = 9.62 − 2 × 9.8 × h
M1
h = 4.70 m (accept 4.7 m)
A1
Part (c)
time of flight = 2 × 0.980 = 1.96 s and horizontal component 12 × 0.6 = 7.2 m s−1; distance = 7.2 × 1.96
M1
= 14.1 m
A1
Exam tip: With tan α = 4/3, draw a 3–4–5 triangle: sin α = 0.8 and cos α = 0.6 exactly, so no angle needs to be found.
A beanbag is thrown from a point on horizontal ground with speed 14 m s−1 at an angle of 60° above the horizontal. The beanbag is modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.
Figure: the beanbag thrown at 60° above the horizontal; not to scale.
(a)
State one modelling assumption, other than treating the beanbag as a particle, that has been made.
(1)
(b)
Show that, 2.0 s after it is thrown, the vertical component of the velocity of the beanbag is 7.48 m s−1 downwards, to 3 significant figures, and verify that it is still in the air.
(2)
(c)
Find the speed of the beanbag 2.0 s after it is thrown.
(2)
(d)
Find the direction of motion of the beanbag at this time.
(2)
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Answer
Mark
Notes
Part (a)
any one: air resistance is negligible / the only force acting is the weight / g is constant
B1
Part (b)
14 sin 60° − 9.8 × 2.0 = 12.12 − 19.6 = −7.48, so 7.48 m s−1 downwards
B1
time of flight 2 × 14 sin 60° ÷ 9.8 = 2.47 s > 2.0 s, so still in the air
B1
Part (c)
horizontal component 14 cos 60° = 7 m s−1; speed = √(72 + 7.482)
M1
= 10.2 m s−1
A1
Part (d)
tan β = 7.48 ÷ 7
M1
β = 46.9° below the horizontal
A1
Exam tip: A direction of motion needs “above” or “below” the horizontal. The sign of the vertical component tells you which.
A particle P is projected from a fixed point O with velocity (6i + 8j) m s−1, where i and j are unit vectors horizontally and vertically upwards respectively. P moves freely under gravity. Take g = 9.8 m s−2.
(a)
Find the speed of projection and the angle of projection.
(3)
(b)
Find the velocity of P 1.5 s after projection, giving your answer in the form ai + bj.
(2)
(c)
Find the position vector of P relative to O at this time.
(2)
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Answer
Mark
Notes
Part (a)
speed = √(62 + 82) = 10 m s−1
B1
tan θ = 8 ÷ 6
M1
θ = 53.1° above the horizontal
A1
Part (b)
v = (6i + 8j) − 9.8 × 1.5j
M1
= 6i − 6.7j m s−1
A1
Part (c)
r = (6i + 8j) × 1.5 − 4.9 × 1.52j
M1
= 9i + 0.975j m
A1
Exam tip: In vector form only the j-component is affected by gravity: v = u − gtj and r = ut − ½gt2j.
Parcels leave the end E of a horizontal conveyor belt with the speed of the belt and fall into an open bin on a horizontal floor. The diagram shows the dimensions: E is 1.6 m above the floor, and the open top of the bin is 0.375 m above the floor and stretches from 0.9 m to 1.5 m horizontally from E. Each parcel is modelled as a particle moving freely under gravity after it leaves E, and the sides of the bin have negligible thickness. A parcel lands in the bin if, when it has fallen to the level of the top of the bin, its horizontal distance from E is between 0.9 m and 1.5 m. Take g = 9.8 m s−2.
(a)
Show that a parcel takes 0.5 s to fall from E to the level of the top of the bin.
(2)
(b)
Find the range of belt speeds for which the parcels land in the bin.
(3)
(c)
The belt is set to 2.2 m s−1. Find the speed of a parcel as it reaches the level of the top of the bin.
(2)
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Answer
Mark
Notes
Part (a)
vertical drop 1.6 − 0.375 = 1.225 m, with zero initial vertical velocity: 1.225 = ½ × 9.8 × t2
M1
t2 = 0.25, so t = 0.5 s (given answer)
A1
Part (b)
horizontal distance = 0.5u, where u m s−1 is the belt speed
A ball is struck from a point O on horizontal ground. It moves freely under gravity and lands on the ground 3 s later. The graph shows how v m s−1, the vertical component of the velocity of the ball (upwards positive), varies with the time t seconds after the ball is struck. The horizontal component of the velocity is constant at 8 m s−1.
(a)
Use the graph to show that the acceleration of the ball is 9.8 m s−2 vertically downwards.
(1)
(b)
Use the graph to find the greatest height of the ball above the ground.
(3)
(c)
Find the speed of projection and the angle of projection.
(2)
(d)
Find the distance from O of the point where the ball lands.
(1)
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Answer
Mark
Notes
Part (a)
gradient = (−14.7 − 14.7) ÷ 3 = −9.8, so the acceleration is 9.8 m s−2 downwards
B1
Part (b)
greatest height when v = 0, at t = 1.5
B1
height = area under the graph from t = 0 to 1.5 = ½ × 1.5 × 14.7
M1
= 11.0 m (11.025; accept 11 m)
A1
Part (c)
speed = √(82 + 14.72) = 16.7 m s−1
B1
angle = tan−1(14.7 ÷ 8) = 61.4° above the horizontal
B1
Part (d)
8 × 3 = 24 m
B1
Exam tip: A velocity–time graph of the vertical motion works exactly as in straight-line kinematics: the gradient is the acceleration and the area is the vertical displacement.
A particle P is projected from a point O on horizontal ground with velocity (ai + bj) m s−1, where i and j are unit vectors horizontally and vertically upwards respectively. P moves freely under gravity and lands at the point L on the ground. The diagram shows the path of P on a grid, with distances in metres. A and B are the positions of P 1 s and 2 s after projection, and their coordinates relative to O are shown. Take g = 9.8 m s−2.
(a)
Use the coordinates of A to find the values of a and b.
(3)
(b)
Find the velocity of P at B, giving your answer in the form pi + qj, and find the speed of P at B.
(3)
(c)
Find the distance OL.
(2)
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Answer
Mark
Notes
Part (a)
horizontally 6 m in 1 s, so a = 6
B1
vertically: b × 1 − 4.9 × 12 = 7.1
M1
b = 12
A1
Part (b)
v = (6i + 12j) − 9.8 × 2j
M1
= 6i − 7.6j m s−1
A1
speed = √(62 + 7.62) = 9.68 m s−1
A1
Part (c)
lands when 12t − 4.9t2 = 0, so t = 12 ÷ 4.9 = 2.449 s
M1
OL = 6 × 2.449 = 14.7 m
A1
Exam tip: Check your values against the second labelled point: with a = 6 and b = 12, the position at t = 2 is (12, 24 − 19.6) = (12, 4.4), which matches B.
Two jets of water, A and B, leave a nozzle N at ground level in a garden. The diagram shows their paths, drawn to scale on a grid. Both jets reach the same greatest height, 3.6 m, and they land on the horizontal ground 6 m and 12 m from N respectively, at the grid points shown. Each drop of water is modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.
(a)
Show that, for both jets, the vertical component of the initial velocity is 8.4 m s−1.
(2)
(b)
Find the time of flight of each jet.
(1)
(c)
Find the speed of projection and the angle of projection of each jet.
(4)
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Answer
Mark
Notes
Part (a)
at the greatest height the vertical component is zero: 0 = uy2 − 2 × 9.8 × 3.6
M1
uy2 = 70.56, so uy = 8.4 (given answer)
A1
Part (b)
2 × 8.4 ÷ 9.8 = 1.71 s for both jets (12/7 s)
B1
Part (c)
horizontal components: distance ÷ time of flight
M1
A: 6 ÷ 1.714 = 3.5 m s−1; B: 12 ÷ 1.714 = 7 m s−1
A1
speeds √(3.52 + 8.42) = 9.1 m s−1 and √(72 + 8.42) = 10.9 m s−1
Exam tip: Equal greatest heights mean equal vertical components, and so equal times of flight. Only the horizontal components differ, in the ratio of the ranges.
A ball is thrown horizontally from an open window. The point of release is 12.5 m vertically above horizontal ground. The ball hits the ground at a point which is 18 m horizontally from the point of release. The ball is modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.
Figure: the ball is thrown horizontally from the window and lands 18 m from the point of release, measured horizontally; not to scale.
(a)
Find the time taken for the ball to reach the ground.
(2)
(b)
Find the speed with which the ball is thrown.
(2)
(c)
Find the speed of the ball as it hits the ground.
(3)
(d)
Find the angle between the direction of motion of the ball and the ground at this instant.
(2)
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Answer
Mark
Notes
Part (a)
12.5 = ½ × 9.8 × t2
M1
t = 1.60 s (1.597)
A1
Part (b)
u = 18 ÷ 1.597
M1
= 11.3 m s−1
A1
Part (c)
vertical component 9.8 × 1.597 = 15.65 m s−1
M1
speed = √(11.272 + 15.652)
M1
= 19.3 m s−1
A1
Part (d)
tan β = 15.65 ÷ 11.27
M1
β = 54.2° (accept 54°)
A1
Exam tip: Keep the unrounded time (1.597… s) in your calculator. Using 1.6 s in later parts can change the third significant figure.
A stone is thrown from a point A on a bridge. A is 22 m vertically above the surface of a river. The stone is thrown with speed 8 m s−1 at an angle of 20° below the horizontal. The stone moves freely under gravity until it hits the water. Take g = 9.8 m s−2.
Figure: the stone thrown from A at 20° below the horizontal; not to scale.
(a)
Find the time taken for the stone to reach the water.
(4)
(b)
Find the horizontal distance travelled by the stone in this time.
(2)
(c)
Explain why only one root of your quadratic equation in part (a) is used.
(1)
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Answer
Mark
Notes
Part (a)
downwards positive: initial vertical component 8 sin 20° = 2.736 m s−1
B1
22 = 2.736t + 4.9t2
M1
solving the quadratic: t = (−2.736 + √(2.7362 + 4 × 4.9 × 22)) ÷ 9.8
M1
t = 1.86 s
A1
Part (b)
8 cos 20° × 1.858
M1
= 14.0 m (accept 14 m)
A1
Part (c)
the other root is negative (−2.42), which would be a time before the stone was thrown
B1
Common error: Using +8 sin 20° as an upward component when the stone is thrown below the horizontal. Draw the velocity first and decide which way the vertical component points.
In a game, a dart is thrown horizontally towards a board. The dart is released at a horizontal distance of 2.4 m from the board, level with the centre of the board, and it hits the board at a point 0.12 m vertically below the centre. The dart is modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.
Figure: the dart's path from the point of release to the board (drop exaggerated); not to scale.
(a)
Find the time taken for the dart to reach the board.
(2)
(b)
Find the speed with which the dart is thrown.
(2)
(c)
The player throws a second dart horizontally from the same point. It hits the board 0.05 m below the centre. Find the speed of this throw.
(3)
(d)
In practice, air resistance acts on the dart. State, with a reason, how this would affect your answer to part (b).
(1)
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Answer
Mark
Notes
Part (a)
0.12 = ½ × 9.8 × t2
M1
t = 0.156 s
A1
Part (b)
u = 2.4 ÷ 0.1565
M1
= 15.3 m s−1
A1
Part (c)
0.05 = 4.9t2
M1
t = 0.101 s
A1
u = 2.4 ÷ 0.101 = 23.8 m s−1
A1
Part (d)
the true speed of throw would need to be greater (the answer is an underestimate), because air resistance slows the dart horizontally so it takes longer to reach the board and drops further
B1
Exam tip: For a horizontal throw, the drop fixes the time (vertical motion only), and the time then fixes the speed (horizontal motion only).
A machine on horizontal ground fires tennis balls with speed 16 m s−1 at an angle θ above the horizontal. The balls are modelled as particles moving freely under gravity, and the balls leave the machine at ground level. Take g = 9.8 m s−2.
(a)
Show that the horizontal distance R m travelled by a ball before it lands is given by R = 256/9.8 sin 2θ.
(4)
(b)
Find the two values of θ for which a ball lands 20 m from the machine.
(3)
(c)
Find the time of flight for the larger value of θ.
(2)
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Answer
Mark
Notes
Part (a)
vertically: 0 = 16 sin θt − 4.9t2
M1
t = 32 sin θ ÷ 9.8 (non-zero root)
A1
R = 16 cos θ × 32 sin θ ÷ 9.8
M1
= 256 × 2 sin θ cos θ ÷ 9.8 = 256/9.8 sin 2θ (given answer, using sin 2θ = 2 sin θ cos θ)
A1
Part (b)
sin 2θ = 20 × 9.8 ÷ 256 = 0.7656
M1
2θ = 49.96° or 130.04°
M1
θ = 25.0° or 65.0°
A1
Part (c)
t = 32 sin 65.0° ÷ 9.8
M1
= 2.96 s
A1
Exam tip: sin 2θ = k has two solutions with 0° < 2θ < 180°. Halving them gives two angles of projection that add to 90°.
A ball is thrown from a point O with speed 13 m s−1 at an angle θ above the horizontal, where tan θ = 12/5. The ball moves freely under gravity in a vertical plane. Take O as the origin, with x m and y m the horizontal and vertically upward displacements of the ball from O. Take g = 9.8 m s−2.
Figure: the ball thrown from O, with the balcony level 5 m above O; tan θ = 12/5; not to scale.
(a)
Show that the equation of the path of the ball is y = 2.4x − 0.196x2.
(4)
(b)
A friend on a balcony catches the ball at a point 5 m above the level of O while the ball is falling. Find the horizontal distance from O to this point.
(3)
(c)
Find the time between the ball first rising through the level of the balcony and being caught.
(2)
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Answer
Mark
Notes
Part (a)
sin θ = 12/13, cos θ = 5/13, so components 5 m s−1 and 12 m s−1
B1
x = 5t, so t = x/5
M1
y = 12(x/5) − 4.9(x/5)2
M1
y = 2.4x − 0.196x2 (given answer)
A1
Part (b)
0.196x2 − 2.4x + 5 = 0
M1
x = (2.4 ± √(2.42 − 4 × 0.196 × 5)) ÷ 0.392 = 2.66 or 9.58
M1
falling, so the larger root: 9.58 m
A1
Part (c)
time = (9.583 − 2.662) ÷ 5
M1
= 1.38 s
A1
Exam tip: The trajectory equation gives positions, not times. Divide a horizontal distance by the constant horizontal velocity to turn it into a time.
A fountain nozzle at ground level projects drops of water with speed 11 m s−1 at an angle of 70° above the horizontal. Each drop is modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.
(a)
Find the horizontal and vertical components of the initial velocity of a drop.
(2)
(b)
Find the two times after projection at which a drop is moving at 45° to the horizontal.
(4)
(c)
Find the height of the drop above the ground at the first of these times.
(2)
(d)
Without further calculation, state the height of the drop at the second time, giving a reason.
(1)
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Answer
Mark
Notes
Part (a)
11 cos 70° = 3.76 m s−1
B1
11 sin 70° = 10.3 m s−1 (10.34)
B1
Part (b)
at 45°, the vertical component has the same size as the horizontal component, 3.762 m s−1
M1
rising: 10.34 − 9.8t = 3.762
M1
t = 0.671 s
A1
falling: 10.34 − 9.8t = −3.762, t = 1.44 s
A1
Part (c)
y = 10.34 × 0.6709 − 4.9 × 0.67092
M1
= 4.73 m
A1
Part (d)
4.73 m, because the two times are equally spaced either side of the time of the top and the path is symmetrical
B1
Exam tip: Moving at 45° means |vvertical| = vhorizontal. There are two such moments: one on the way up and one on the way down.
A ball is kicked from a point O on horizontal ground with speed u m s−1 at an angle θ above the horizontal, and moves freely under gravity. The graph shows the path of the ball, where x m and y m are its horizontal and vertical distances from O. The highest point of the path is at a grid point, and the ball lands at x = 20. A vertical wall of height 3 m stands at x = 16. Take g = 9.8 m s−2.
(a)
Write down the greatest height of the ball and its horizontal distance from O at that moment. Hence show that tan θ = 1.
(3)
(b)
Find the value of u.
(2)
(c)
Show that the equation of the path is y = x − 0.05x2.
(2)
(d)
Determine whether the ball passes over the wall and, if so, by how much it clears it.
(2)
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Answer
Mark
Notes
Part (a)
greatest height 5 m, at x = 10
B1
if t1 is the time to the top: 5 = ½ u sin θt1 and 10 = u cos θt1
M1
dividing: ½ tan θ = ½, so tan θ = 1 (given answer)
A1
Part (b)
(u sin 45°)2 = 2 × 9.8 × 5 = 98
M1
u2 = 196, u = 14
A1
Part (c)
y = x tan 45° − 9.8x2 ÷ (2 × 142 × cos2 45°)
M1
= x − 9.8x2 ÷ 196 = x − 0.05x2 (given answer)
A1
Part (d)
at x = 16: y = 16 − 0.05 × 256 = 3.2
M1
3.2 > 3, so it clears the wall by 0.2 m
A1
Exam tip: Use the graph for values at grid points only. A clearance of 0.2 m is too small to read from a diagram, so calculate it from the equation of the path.
A ball is kicked from a point O on horizontal ground with speed u m s−1 at 30° above the horizontal. It moves freely under gravity and lands at the point T on the flat top of a platform. The diagram shows the dimensions: the platform is 1.5 m high, its front edge is 15 m horizontally from O, and T is 18 m horizontally from O. Take g = 9.8 m s−2.
(a)
Find the value of u.
(4)
(b)
Find the time taken for the ball to reach T.
(2)
(c)
Show that the ball is moving downwards when it reaches T.
(2)
(d)
Verify that the ball passes above the front edge of the platform.
(1)
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Answer
Mark
Notes
Part (a)
using y = x tan θ − gx2 ÷ (2u2 cos2θ) or eliminating t
M1
1.5 = 18 tan 30° − 9.8 × 182 ÷ (2u2 × 0.75)
A1
u2 = 3175.2 ÷ (1.5 × (10.392 − 1.5)) = 238.0
M1
u = 15.4
A1
Part (b)
t = 18 ÷ (15.43 cos 30°)
M1
= 1.35 s
A1
Part (c)
vertical component = 15.43 sin 30° − 9.8 × 1.347
M1
= −5.49 < 0, so the ball is moving downwards
A1
Part (d)
at x = 15: y = 15 tan 30° − 9.8 × 225 ÷ (2 × 238.0 × 0.75) = 2.49 > 1.5, so it passes above the edge
B1
Exam tip: When the speed is the unknown, the equation of the path, y = x tan θ − gx2/(2u2 cos2 θ), gives u2 directly without finding the time first.
A ball is projected from a point O on horizontal ground and moves freely under gravity. The diagram shows its path. At the point Q, 1.5 s after projection, the ball is moving with speed 10 m s−1 in the direction shown, 30° below the horizontal. Take g = 9.8 m s−2.
(a)
Find the horizontal and vertical components of the initial velocity of the ball.
(3)
(b)
Find the speed and the angle of projection.
(3)
(c)
Find the height of Q above the ground.
(2)
(d)
Find how long after passing through Q the ball lands.
(1)
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Answer
Mark
Notes
Part (a)
horizontal component = 10 cos 30° = 8.66 m s−1 (5√3)
B1
vertically, upwards positive: −10 sin 30° = uy − 9.8 × 1.5
M1
uy = −5 + 14.7 = 9.7 m s−1
A1
Part (b)
speed = √(8.662 + 9.72)
M1
= 13.0 m s−1
A1
angle = tan−1(9.7 ÷ 8.66) = 48.2° above the horizontal
A1
Part (c)
9.7 × 1.5 − 4.9 × 1.52
M1
= 3.525 m (accept 3.52 or 3.53)
A1
Part (d)
time of flight 2 × 9.7 ÷ 9.8 = 1.980 s, so 1.980 − 1.5 = 0.480 s
B1
Common error: Taking the vertical component at Q as +5 m s−1. The ball is moving 30° below the horizontal, so with upwards positive the component is −5 m s−1.
A ball is thrown from a point A, 2.1 m above horizontal ground, and moves freely under gravity until it hits the ground. The graphs show x m, the horizontal displacement of the ball from A, and y m, the height of the ball above the ground, against the time t seconds after it is thrown. The labelled points give exact values. Take g = 9.8 m s−2.
(a)
Use the graphs to find the horizontal and vertical components of the initial velocity of the ball.
(3)
(b)
Find the speed and the angle of projection.
(2)
(c)
Find the time at which the ball hits the ground.
(3)
(d)
Find the horizontal distance travelled by the ball before it hits the ground.
(1)
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Answer
Mark
Notes
Part (a)
horizontal component = 15 ÷ 2 = 7.5 m s−1
B1
the highest point is at t = 1.0, so 0 = uy − 9.8 × 1.0 (or 7.0 − 2.1 = uy2 ÷ 19.6)
M1
uy = 9.8 m s−1
A1
Part (b)
speed = √(7.52 + 9.82) = 12.3 m s−1
B1
angle = tan−1(9.8 ÷ 7.5) = 52.6° above the horizontal
Exam tip: Read values only at labelled points or grid crossings. The landing time is not on a gridline here, so it has to be calculated from the vertical motion.
A ball is thrown from the edge E of the flat roof of building A with speed 12 m s−1 at 25° above the horizontal. The diagram shows the dimensions: E is 20 m above horizontal ground, and building B has a flat roof 8 m above the ground that begins 10 m horizontally from E and is 15 m long. The ball moves freely under gravity in the vertical plane shown. Take g = 9.8 m s−2.
(a)
Find the time taken for the ball to fall to the level of the roof of B.
(4)
(b)
Show that the ball lands on the roof of B, and find its distance from the near edge of that roof.
(3)
(c)
Find the speed of the ball as it lands.
(3)
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Answer
Mark
Notes
Part (a)
vertical component of initial velocity 12 sin 25° = 5.071 m s−1
B1
displacement 8 − 20 = −12 m: −12 = 5.071t − 4.9t2
M1
t = (5.071 + √(5.0712 + 4 × 4.9 × 12)) ÷ 9.8
M1
t = 2.17 s (2.166)
A1
Part (b)
horizontal distance = 12 cos 25° × 2.166
M1
= 23.6 m
A1
10 < 23.6 < 25, so it lands on the roof of B, 13.6 m from the near edge
Exam tip: Measure every vertical displacement from the point of projection. Landing on a roof 8 m high, starting from 20 m, is a displacement of −12 m.
A stunt rider leaves the lip L of a ramp with speed u m s−1 at 20° above the horizontal. The diagram shows the landing deck, whose horizontal top is level with L. The near end of the deck is 9 m horizontally from L and the deck is 6 m long. The rider and bike are modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.
(a)
Find the range of values of u for which the rider lands on the deck.
(5)
(b)
Given that u = 13, find the greatest height of the rider above the level of L.
(2)
(c)
Given that u = 13, find the distance from the near end of the deck to the point where the rider lands.
(2)
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Answer
Mark
Notes
Part (a)
time to return to the level of L: t = 2u sin 20° ÷ 9.8
M1
horizontal distance = u cos 20° × 2u sin 20° ÷ 9.8 = u2 sin 40° ÷ 9.8
A1
9 ≤ u2 sin 40° ÷ 9.8 ≤ 15
M1
lower limit u ≥ 11.7
A1
upper limit u ≤ 15.1, so 11.7 ≤ u ≤ 15.1
A1
Part (b)
(13 sin 20°)2 ÷ (2 × 9.8)
M1
= 1.01 m
A1
Part (c)
132 sin 40° ÷ 9.8 = 11.08 m from L
M1
11.08 − 9 = 2.08 m (accept 2.1 m)
A1
Common error: Using only the 9 m gap. “Lands on the deck” has a far end too, so there is an upper limit on the speed as well as a lower one.
A water balloon is launched from a point O with speed 24 m s−1 at an angle θ above the horizontal. It passes through an open window W. The point W is 36 m horizontally from O and 8.5 m vertically above the level of O. The balloon is modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.
Figure: the launch point O and the window W; the path of the balloon is not shown; not to scale.
(a)
Show that 441 tan2θ − 1440 tan θ + 781 = 0.
(5)
(b)
Hence find the two possible values of θ.
(3)
(c)
For the smaller value of θ, find the time taken for the balloon to reach W.
(2)
(d)
For the smaller value of θ, determine whether the balloon is moving upwards or downwards as it passes through W.
(2)
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Answer
Mark
Notes
Part (a)
horizontally: 36 = 24 cos θ × t, so t = 1.5/cos θ
M1
vertically: 8.5 = 24 sin θ × t − 4.9t2
M1
8.5 = 36 tan θ − 11.025 sec2θ
A1
use of sec2θ = 1 + tan2θ: 8.5 = 36 tan θ − 11.025 − 11.025 tan2θ
A particle P is projected from a point A with velocity (8i + 10j) m s−1, where i and j are unit vectors horizontally and vertically upwards respectively. The point O is on horizontal ground and the position vector of A relative to O is (2i + 3j) m. P moves freely under gravity until it hits the ground. Take g = 9.8 m s−2.
Figure: P is projected from A, whose position relative to O is (2i + 3j) m; not to scale.
(a)
Find the position vector of P relative to O at time t seconds after projection.
(2)
(b)
Find the time taken for P to reach the ground.
(3)
(c)
Find the distance from O of the point where P hits the ground.
(2)
(d)
Find the greatest height of P above the ground, and its speed at that instant.
(3)
(e)
Find the time at which P is moving in a direction perpendicular to its initial direction of motion.
(3)
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Answer
Mark
Notes
Part (a)
r = (2i + 3j) + (8i + 10j)t − 4.9t2j
M1
r = (2 + 8t)i + (3 + 10t − 4.9t2)j
A1
Part (b)
3 + 10t − 4.9t2 = 0
M1
t = (10 + √(100 + 58.8)) ÷ 9.8
M1
= 2.31 s (2.306)
A1
Part (c)
2 + 8 × 2.306
M1
= 20.5 m
A1
Part (d)
at the top the j-component of velocity is zero: 0 = 102 − 2 × 9.8 × s, s = 5.10
M1
height = 3 + 5.10 = 8.10 m
A1
speed = 8 m s−1 (the i-component only)
B1
Part (e)
v = 8i + (10 − 9.8t)j; perpendicular to 8i + 10j when the scalar product is zero, or gradient −0.8
M1
64 + 10(10 − 9.8t) = 0
M1
t = 164 ÷ 98 = 1.67 s, which is before P lands
A1
Exam tip: If the particle does not start at O, add the starting position vector: r = r0 + ut − ½gt2j.
A stone is catapulted from a point C on the top of a vertical castle wall. C is 30 m above horizontal ground. The stone is projected with speed 24 m s−1 at an angle of 40° above the horizontal and moves freely under gravity until it hits the ground. Take g = 9.8 m s−2.
Figure: the stone catapulted from C on top of the wall; not to scale.
(a)
Find the greatest height of the stone above the ground.
(3)
(b)
Find the time taken for the stone to reach the ground.
(4)
(c)
Find the horizontal distance from the foot of the wall to the point where the stone lands.
(2)
(d)
Find the length of time for which the stone is more than 35 m above the ground.
(4)
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Answer
Mark
Notes
Part (a)
vertical component 24 sin 40° = 15.43 m s−1; 0 = 15.432 − 2 × 9.8 × s
M1
s = 12.14 m
A1
height above ground = 30 + 12.14 = 42.1 m
A1
Part (b)
upwards positive, displacement −30
B1
−30 = 15.43t − 4.9t2
M1
t = (15.43 + √(15.432 + 4 × 4.9 × 30)) ÷ 9.8
M1
= 4.51 s (4.507)
A1
Part (c)
24 cos 40° × 4.507 = 18.39 × 4.507
M1
= 82.9 m
A1
Part (d)
35 m above ground is 5 m above C: 5 = 15.43t − 4.9t2
Common error: Using the level-ground formula for the range when the stone lands below its launch point. Find the time from the vertical quadratic first, then use the horizontal motion.
A particle is projected from a point O on horizontal ground with speed u m s−1 at an angle θ above the horizontal. It moves freely under gravity, where the acceleration due to gravity is g m s−2.
(a)
Show that the time taken for the particle to return to the ground is 2u sin θ/g.
(3)
(b)
Hence show that the horizontal distance travelled in this time is u2 sin 2θ/g.
(2)
(c)
A long-jumper takes off with speed 8.5 m s−1 and is modelled as a particle moving freely under gravity, taking off and landing on the same horizontal level. Using g = 9.8, find the two possible angles of take-off that give a jump of 6.4 m.
(3)
(d)
Find the greatest height above the take-off level reached by the long-jumper for the smaller of these angles.
(3)
(e)
State one way in which the particle model is unrealistic for a long-jumper.
(1)
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Answer
Mark
Notes
Part (a)
vertically: 0 = (u sin θ)t − ½gt2
M1
t(u sin θ − ½gt) = 0
M1
t ≠ 0, so t = 2u sin θ/g (given answer)
A1
Part (b)
x = u cos θ × 2u sin θ/g
M1
= u2(2 sin θ cos θ)/g = u2 sin 2θ/g (given answer)
A1
Part (c)
6.4 = 8.52 sin 2θ ÷ 9.8
M1
sin 2θ = 0.8681, 2θ = 60.24° or 119.76°
M1
θ = 30.1° or 59.9°
A1
Part (d)
vertical component 8.5 sin 30.12° = 4.265 m s−1
M1
0 = 4.2652 − 2 × 9.8 × h
M1
h = 0.928 m (accept 0.93 m)
A1
Part (e)
any one: the jumper has size and changes body shape, so cannot be a single point / the centre of mass does not start and end at ground level / air resistance acts
B1
Exam tip: In a “show that” with a given answer, every step must be written down, including the identity sin 2θ = 2 sin θ cos θ.
A ball is projected from a point O on horizontal ground with speed 15 m s−1 at an angle θ above the horizontal, and moves freely under gravity. The diagram shows a vertical post standing on the ground; its top P is 12 m horizontally from O and 4 m above the ground. The ball must pass above P. Take g = 9.8 m s−2.
(a)
Show that the ball passes above P when 98 tan2θ − 375 tan θ + 223 < 0.
(5)
(b)
Hence find the range of values of θ for which the ball passes above P.
(4)
(c)
For θ = 45°, find the height by which the ball clears P and the distance beyond the post at which the ball lands.
(3)
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Answer
Mark
Notes
Part (a)
horizontally 12 = 15 cos θt, so t = 12 ÷ (15 cos θ)
M1
height at the post: y = 12 tan θ − 4.9 × 144 ÷ (225 cos2θ) = 12 tan θ − 3.136 sec2θ
The diagram shows a point O on horizontal ground and a vertical tower whose foot F is 20 m from O; the top T of the tower is 15 m above F. At time t = 0, a particle A is projected from O with velocity (10i + 15j) m s−1, and at the same instant a particle B is projected horizontally from T towards O with speed V m s−1. Here i and j are unit vectors horizontally (in the direction from O to F) and vertically upwards. Both particles move freely under gravity in the vertical plane through O, F and T. Take g = 9.8 m s−2.
(a)
Write down the position vectors of A and of B, relative to O, at time t seconds.
(3)
(b)
Given that A and B collide, show that they collide 1 s after projection, and find V.
(3)
(c)
Find the position vector of the point where they collide.
(1)
(d)
Find the velocities of A and B just before they collide, and the angle between their directions of motion at that instant.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
rA = 10ti + (15t − 4.9t2)j
B1
rB = (20 − Vt)i …
B1
… + (15 − 4.9t2)j
B1
Part (b)
equating j-components: 15t − 4.9t2 = 15 − 4.9t2
M1
15t = 15, so t = 1 (given answer)
A1
equating i-components at t = 1: 10 = 20 − V, V = 10
A1
Part (c)
10i + 10.1j m
B1
Part (d)
vA = 10i + (15 − 9.8)j = 10i + 5.2j m s−1
B1
vB = −10i − 9.8j m s−1
B1
cos φ = (−100 − 50.96) ÷ (11.27 × 14.00)
M1
φ = 163°
A1
Exam tip: Two particles collide when both components of their position vectors are equal at the same time. Here the gravity terms cancel in the j-components, so t comes out without knowing V.
A ball is projected from a point A at the top of a vertical cliff with speed u m s−1 at an angle α above the horizontal, where tan α = 3/4. The ball moves freely under gravity and hits the sea at the point B, 5 s after projection. The diagram shows that A is h m above the sea and that B is 60 m horizontally from A. Take g = 9.8 m s−2.
(a)
Find the value of u.
(2)
(b)
Find the value of h.
(3)
(c)
Find the speed of the ball as it hits the sea, and its direction of motion at that instant.
(4)
(d)
Find the length of time for which the speed of the ball is less than 13 m s−1.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
cos α = 0.8: u × 0.8 × 5 = 60
M1
u = 15
A1
Part (b)
vertical component 15 × 0.6 = 9 m s−1
B1
−h = 9 × 5 − 4.9 × 52
M1
h = 77.5
A1
Part (c)
vertical component 9 − 9.8 × 5 = −40 m s−1
M1
speed = √(122 + 402) = 41.8 m s−1
A1
tan β = 40 ÷ 12
M1
β = 73.3° below the horizontal
A1
Part (d)
speed2 = 122 + vy2 < 132
M1
|vy| < 5
A1
−5 < 9 − 9.8t < 5, so 0.408 < t < 1.43
M1
length of time = 10 ÷ 9.8 = 1.02 s
A1
Exam tip: The horizontal component never changes, so the speed is least at the top of the path (12 m s−1 here). A condition on the speed is really a condition on the vertical component.
In a sports hall, a ball is launched from a point O on the floor with speed 21 m s−1 at an angle θ above the horizontal. It moves freely under gravity and is aimed at the point P on the edge of a balcony. The diagram shows that P is 30 m horizontally from O and 5 m above the floor, and that the horizontal ceiling is 12 m above the floor. Take g = 9.8 m s−2.
(a)
Show that, if the ball passes through P, then 2 tan2θ − 6 tan θ + 3 = 0.
(5)
(b)
Find the exact values of tan θ, and hence the two possible angles of projection.
(3)
(c)
By finding the greatest height of the ball for each angle, show that only one of these angles can be used.
(3)
(d)
For the angle that can be used, find the time taken for the ball to reach P.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
horizontally t = 30 ÷ (21 cos θ), substituted into the vertical equation
M1
5 = 30 tan θ − 4.9 × 900 ÷ (441 cos2θ)
A1
4.9 × 900 ÷ 441 = 10 and sec2θ = 1 + tan2θ
M1
5 = 30 tan θ − 10 − 10 tan2θ
A1
10 tan2θ − 30 tan θ + 15 = 0, so 2 tan2θ − 6 tan θ + 3 = 0 (given answer)
A1
Part (b)
tan θ = (6 ± √12) ÷ 4
M1
= (3 ± √3)/2
A1
θ = 32.4° or 67.1°
A1
Part (c)
greatest height = (21 sin θ)2 ÷ (2 × 9.8)
M1
for 32.4°: 6.45 m; for 67.1°: 19.1 m
A1
19.1 > 12 and that path reaches its top (16.1 m from O) before P, so it would hit the ceiling; only 32.4° can be used (6.45 < 12)
A1
Part (d)
t = 30 ÷ (21 cos 32.37°)
M1
= 1.69 s
A1
Common error: Giving both roots of the tan θ quadratic as answers. Each root is a possible path through P; check each one against the setting, here the ceiling.
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