Exam-style questionsProjectiles questions with mark schemes
Easiest first. Try each question on paper, then open its mark scheme: M marks are for method, A marks for accurate answers that follow from the method, B marks for independent results.
Question 1easy7 marksYear 13
A stone is thrown horizontally with speed 12 m s\(^{-1}\) from the top of a vertical cliff, 45 m above the sea. The stone is modelled as a particle moving freely under gravity.
(a) Find the time taken for the stone to reach the sea. (2)
(b) Find the horizontal distance travelled by the stone. (2)
(c) Find the speed of the stone as it reaches the sea. (3)
Mark scheme and answer
(a) Vertical: \(45 = \tfrac12(9.8)t^2\) M1
\(t = 3.03\) s A1
(b) \(12\times3.030 = 36.4\) m M1A1
(c) \(v_y = 9.8\times3.030 = 29.70\) M1
\(\sqrt{12^2 + 29.70^2} = 32.0\) m s\(^{-1}\) M1A1
Answer: (a) 3.03 s; (b) 36.4 m; (c) 32.0 m s⁻¹
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)). All three boards use g = 9.8 m s⁻² unless told otherwise; answers using g = 9.8 are usually given to 2 or 3 s.f. AQA sometimes asks candidates to "show that" the trajectory equation; OCR often uses i, j vector notation for projectiles.
Question 2easy7 marksYear 13
A ball is kicked from a point \(O\) on horizontal ground with speed 20 m s\(^{-1}\) at an angle of 35° above the horizontal. The ball is modelled as a particle and air resistance is ignored.
(a) Find the greatest height of the ball above the ground. (3)
(b) Find the time taken for the ball to return to the ground. (2)
(c) Find the horizontal distance from \(O\) to the point where the ball lands. (2)
Mark scheme and answer
(a) \(0 = (20\sin35°)^2 - 2(9.8)H\) M1A1
\(H = 6.71\) m A1
(b) \(0 = 20\sin35°\,t - 4.9t^2\) M1
\(t = 2.34\) s A1
(c) \(20\cos35°\times2.341\) M1
\(= 38.4\) m A1
Answer: (a) 6.71 m; (b) 2.34 s; (c) 38.4 m
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)). All three boards use g = 9.8 m s⁻² unless told otherwise; answers using g = 9.8 are usually given to 2 or 3 s.f. AQA sometimes asks candidates to "show that" the trajectory equation; OCR often uses i, j vector notation for projectiles.
Question 3easy7 marksYear 13
A shot-putter releases a shot 1.8 m above horizontal ground with speed 14 m s\(^{-1}\) at 25° above the horizontal.
(a) Find the horizontal distance travelled by the shot before it hits the ground. (5)
(b) State two modelling assumptions you have made. (2)
Mark scheme and answer
(a) \(-1.8 = 14\sin25°\,t - 4.9t^2\) M1A1
\(t = 1.46\) A1
\(x = 14\cos25°\times1.459 = 18.5\) m M1A1
(b) Shot is a particle B1
Air resistance is negligible B1
Answer: (a) 18.5 m; (b) particle; no air resistance
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)). All three boards use g = 9.8 m s⁻² unless told otherwise; answers using g = 9.8 are usually given to 2 or 3 s.f. AQA sometimes asks candidates to "show that" the trajectory equation; OCR often uses i, j vector notation for projectiles.
Question 4medium6 marksYear 13
A ball is thrown from a point 2 m above horizontal ground with speed 25 m s\(^{-1}\) at an angle \(\alpha\) above the horizontal, where \(\tan\alpha = \tfrac34\).
(a) Write down the initial horizontal and vertical components of the velocity. (1)
(b) Find the time taken for the ball to hit the ground. (4)
(c) Find the horizontal distance travelled by the ball. (1)
Mark scheme and answer
(a) \(\cos\alpha = \tfrac45,\ \sin\alpha = \tfrac35\): horizontal 20 m s\(^{-1}\), vertical 15 m s\(^{-1}\) B1
(b) \(-2 = 15t - 4.9t^2\) M1A1
\(4.9t^2 - 15t - 2 = 0\), \(t = 3.19\) s dM1A1
(c) \(20\times3.189 = 63.8\) m B1ft
Answer: (a) 20 and 15 m s⁻¹; (b) 3.19 s; (c) 63.8 m
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)). All three boards use g = 9.8 m s⁻² unless told otherwise; answers using g = 9.8 are usually given to 2 or 3 s.f. AQA sometimes asks candidates to "show that" the trajectory equation; OCR often uses i, j vector notation for projectiles.
Question 5medium6 marksYear 13
A cricket ball is struck from ground level with speed 22 m s\(^{-1}\) at 40° above the horizontal towards a vertical fence 4 m high, 30 m away on horizontal ground.
(a) Show that the ball clears the fence, and find by how much. (5)
(b) State one modelling assumption and explain how it affects your answer. (1)
Mark scheme and answer
(a) Time to reach fence: \(t = \dfrac{30}{22\cos40°} = 1.780\) s M1A1
Height: \(22\sin40°\,t - 4.9t^2 = 9.65\) m M1A1
\(9.65 \gt 4\), clears by 5.6 m A1
(b) Air resistance ignored – in reality the ball would slow and be lower, so it may clear the fence by less B1
Answer: (a) height 9.65 m at fence; clears by 5.6 m; (b) air resistance would reduce the height
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)). All three boards use g = 9.8 m s⁻² unless told otherwise; answers using g = 9.8 are usually given to 2 or 3 s.f. AQA sometimes asks candidates to "show that" the trajectory equation; OCR often uses i, j vector notation for projectiles.
Question 6medium5 marksYear 13
A particle \(P\) is projected from the origin \(O\) with velocity \((8\mathbf{i} + 14\mathbf{j})\) m s\(^{-1}\), where \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal and vertically upward unit vectors. \(P\) moves freely under gravity.
(a) Find the position vector of \(P\) at time \(t = 2\) seconds. (2)
(b) Find the time at which \(P\) is moving at 45° below the horizontal. (3)
Mark scheme and answer
(a) \(\mathbf{r} = (8\mathbf{i} + 14\mathbf{j})t - 4.9t^2\mathbf{j}\) M1
\(\mathbf{r} = (16\mathbf{i} + 8.4\mathbf{j})\) m A1
(b) \(\mathbf{v} = 8\mathbf{i} + (14 - 9.8t)\mathbf{j}\) M1
\(14 - 9.8t = -8\) M1
\(t = 2.24\) s A1
Answer: (a) (16i + 8.4j) m; (b) 2.24 s
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)). All three boards use g = 9.8 m s⁻² unless told otherwise; answers using g = 9.8 are usually given to 2 or 3 s.f. AQA sometimes asks candidates to "show that" the trajectory equation; OCR often uses i, j vector notation for projectiles.
Question 7medium5 marksYear 13
A golf ball is struck from a point on horizontal ground at 30° above the horizontal and lands 60 m away on the same horizontal level.
(a) Find the speed with which the ball was struck. (4)
(b) Find the time of flight. (1)
Mark scheme and answer
(a) \(t = \dfrac{2u\sin30°}{g} = \dfrac{u}{9.8}\) M1
\(60 = u\cos30°\times\dfrac{u}{9.8}\) M1A1
\(u = 26.1\) m s\(^{-1}\) A1
(b) \(2.66\) s B1ft
Answer: (a) 26.1 m s⁻¹; (b) 2.66 s
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)). All three boards use g = 9.8 m s⁻² unless told otherwise; answers using g = 9.8 are usually given to 2 or 3 s.f. AQA sometimes asks candidates to "show that" the trajectory equation; OCR often uses i, j vector notation for projectiles.
Question 8medium5 marksYear 13
A ball is thrown from a window 30 m above horizontal ground with speed 10 m s\(^{-1}\) at 15° below the horizontal.
(a) Find the time taken for the ball to reach the ground. (3)
(b) Find the horizontal distance travelled. (2)
Mark scheme and answer
(a) Taking down as positive: \(30 = 10\sin15°\,t + 4.9t^2\) M1A1
\(t = 2.22\) s A1
(b) \(10\cos15°\times2.224 = 21.5\) m M1A1
Answer: (a) 2.22 s; (b) 21.5 m
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)). All three boards use g = 9.8 m s⁻² unless told otherwise; answers using g = 9.8 are usually given to 2 or 3 s.f. AQA sometimes asks candidates to "show that" the trajectory equation; OCR often uses i, j vector notation for projectiles.
Question 9medium6 marksYear 13
A particle is projected from ground level with speed 30 m s\(^{-1}\) at 60° above the horizontal.
(a) Find the speed of the particle when it is 20 m above the ground. (5)
(b) Explain why the particle has this speed at two different times. (1)
Mark scheme and answer
(a) \(v_y^2 = (30\sin60°)^2 - 2(9.8)(20) = 283.0\) M1A1
\(v_x = 15\) B1
speed \(= \sqrt{225 + 283.0} = 22.5\) m s\(^{-1}\) M1A1
(b) It is at 20 m once on the way up and once on the way down; the vertical components are equal in magnitude, opposite in direction B1
Answer: (a) 22.5 m s⁻¹; (b) up and down
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)). All three boards use g = 9.8 m s⁻² unless told otherwise; answers using g = 9.8 are usually given to 2 or 3 s.f. AQA sometimes asks candidates to "show that" the trajectory equation; OCR often uses i, j vector notation for projectiles.
Question 10medium6 marksYear 13
A javelin is modelled as a particle thrown from ground level with speed 18 m s\(^{-1}\) at 40° above the horizontal.
(a) Find the direction of motion of the javelin 2.2 seconds after it is thrown. (5)
(b) Explain why modelling the javelin as a particle may be unrealistic. (1)
Mark scheme and answer
(a) \(v_x = 18\cos40° = 13.79\) B1
\(v_y = 18\sin40° - 9.8(2.2) = -9.990\) M1A1
\(\tan\theta = \dfrac{9.990}{13.79}\), \(\theta = 35.9°\) below the horizontal M1A1
(b) A javelin is long and its orientation/aerodynamic lift affects its flight B1
Answer: (a) 35.9° below horizontal; (b) shape/lift ignored
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)). All three boards use g = 9.8 m s⁻² unless told otherwise; answers using g = 9.8 are usually given to 2 or 3 s.f. AQA sometimes asks candidates to "show that" the trajectory equation; OCR often uses i, j vector notation for projectiles.
Question 11medium6 marksYear 13
A parcel is dropped from a low-flying drone moving horizontally at constant speed \(u\) m s\(^{-1}\) at a height of 20 m. The parcel hits the ground moving at 50° below the horizontal.
(a) Find the vertical component of the parcel's velocity on impact. (2)
(b) Find \(u\). (2)
(c) Find the horizontal distance between the point of release and the point of impact. (2)
Mark scheme and answer
(a) \(v_y^2 = 2(9.8)(20)\), \(v_y = 19.8\) m s\(^{-1}\) M1A1
(b) \(\tan50° = \dfrac{19.80}{u}\) M1
\(u = 16.6\) A1
(c) \(t = 2.020\) s; \(x = 33.6\) m M1A1
Answer: (a) 19.8 m s⁻¹; (b) u = 16.6; (c) 33.6 m
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)). All three boards use g = 9.8 m s⁻² unless told otherwise; answers using g = 9.8 are usually given to 2 or 3 s.f. AQA sometimes asks candidates to "show that" the trajectory equation; OCR often uses i, j vector notation for projectiles.
Question 12hard6 marksYear 13
A particle is projected from a point \(O\) on horizontal ground with speed 18 m s\(^{-1}\) at an angle \(\theta\) above the horizontal. It lands at a point on the ground 25 m from \(O\).
(a) Show that \(\sin2\theta = \dfrac{25g}{324}\). (3)
(b) Find the two possible values of \(\theta\). (3)
Mark scheme and answer
(a) Time of flight \(t = \dfrac{2\times18\sin\theta}{g}\) M1
\(25 = 18\cos\theta\times\dfrac{36\sin\theta}{g}\) M1
\(25g = 324\times2\sin\theta\cos\theta = 324\sin2\theta\) A1
(b) \(\sin2\theta = 0.7562\) M1
\(\theta = 24.6°\) or \(\theta = 65.4°\) A1A1
Answer: (a) shown; (b) 24.6° or 65.4°
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)). All three boards use g = 9.8 m s⁻² unless told otherwise; answers using g = 9.8 are usually given to 2 or 3 s.f. AQA sometimes asks candidates to "show that" the trajectory equation; OCR often uses i, j vector notation for projectiles.
Question 13hard6 marksYear 13
A particle is projected from the origin \(O\) with speed \(u\) m s\(^{-1}\) at an angle \(\alpha\) above the horizontal. At time \(t\) seconds its horizontal and vertical displacements from \(O\) are \(x\) m and \(y\) m.
(a) Show that \(y = x\tan\alpha - \dfrac{gx^2}{2u^2}\left(1 + \tan^2\alpha\right)\). (4)
(b) Given \(u = 15\) and \(\alpha = 50°\), find the height of the particle when \(x = 12\). (2)
Mark scheme and answer
(a) \(x = u\cos\alpha\, t\), \(y = u\sin\alpha\, t - \tfrac12gt^2\) M1
Substitute \(t = \dfrac{x}{u\cos\alpha}\) M1
\(y = x\tan\alpha - \dfrac{gx^2}{2u^2\cos^2\alpha}\) A1
Use \(\sec^2\alpha = 1 + \tan^2\alpha\) to obtain the given result A1
(b) \(y = 12\tan50° - \dfrac{9.8\times144}{2\times225}(1+\tan^250°)\) M1
\(= 6.71\) m A1
Answer: (a) shown; (b) 6.71 m
Board notes: Edexcel and AQA both set "show that" trajectory derivations; OCR more often asks for the Cartesian equation directly. Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)).
Question 14hard7 marksYear 13
A particle is projected from a point on horizontal ground with speed \(U\) at angle \(\alpha\) above the horizontal.
(a) Show that the horizontal range is \(R = \dfrac{U^2\sin2\alpha}{g}\). (4)
(b) Find the greatest possible range for a projectile launched from ground level at 16 m s\(^{-1}\), and the angle at which it occurs. (2)
(c) Give one reason why a real object projected in this way would travel less far than this. (1)
Mark scheme and answer
(a) Vertical: \(0 = U\sin\alpha\,t - \tfrac12gt^2 \Rightarrow t = \dfrac{2U\sin\alpha}{g}\) M1A1
\(R = U\cos\alpha\times\dfrac{2U\sin\alpha}{g} = \dfrac{U^2\sin2\alpha}{g}\) M1A1
(b) Max when \(\sin2\alpha = 1\), \(\alpha = 45°\) B1
\(R = \dfrac{256}{9.8} = 26.1\) m B1
(c) Air resistance would slow it / the object is not a particle (its size, shape or spin affect the flight) B1
Answer: (a) shown; (b) 26.1 m at 45°; (c) air resistance / not a particle
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)). All three boards use g = 9.8 m s⁻² unless told otherwise; answers using g = 9.8 are usually given to 2 or 3 s.f. AQA sometimes asks candidates to "show that" the trajectory equation; OCR often uses i, j vector notation for projectiles.
Question 15hard5 marksYear 13
A firework rocket, after its fuel is spent, is modelled as a particle projected from ground level with speed 24 m s\(^{-1}\) at 55° above the horizontal.
(a) Find the length of time for which the particle is more than 15 m above the ground. (5)
Mark scheme and answer
(a) \(24\sin55°\,t - 4.9t^2 = 15\) M1A1
\(t = 1.02\) or \(2.99\) dM1A1
Time \(= 1.96\) s A1
Answer: 1.96 s
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)). All three boards use g = 9.8 m s⁻² unless told otherwise; answers using g = 9.8 are usually given to 2 or 3 s.f. AQA sometimes asks candidates to "show that" the trajectory equation; OCR often uses i, j vector notation for projectiles.
Question 16hard6 marksYear 13
A golfer strikes a ball from a point \(O\) with speed 30 m s\(^{-1}\) at 50° above the horizontal. The ball lands, on its way down, on a green that is horizontal and 6 m higher than \(O\).
(a) Find the time of flight. (4)
(b) Find the horizontal distance from \(O\) to where the ball lands. (2)
Mark scheme and answer
(a) \(6 = 30\sin50°\,t - 4.9t^2\) M1A1
\(t = 0.278\) or \(t = 4.41\) dM1
Landing on way down: \(t = 4.41\) s A1
(b) \(30\cos50°\times4.413 = 85.1\) m M1A1
Answer: (a) 4.41 s; (b) 85.1 m
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)). All three boards use g = 9.8 m s⁻² unless told otherwise; answers using g = 9.8 are usually given to 2 or 3 s.f. AQA sometimes asks candidates to "show that" the trajectory equation; OCR often uses i, j vector notation for projectiles.
Question 17hard4 marksYear 13
In a fairground game, a small ball rolls off a horizontal table top 1.2 m above the floor, leaving the edge horizontally with speed \(u\) m s\(^{-1}\). To win, the ball must first hit the floor at a horizontal distance of between 2.0 m and 2.5 m from the edge of the table.
(a) Find the range of values of \(u\) for which the player wins. (4)
Mark scheme and answer
(a) \(1.2 = 4.9t^2\), \(t = 0.4949\) s M1A1
\(2.0 \le u\times0.4949 \le 2.5\) M1
\(4.04 \le u \le 5.05\) A1
Answer: 4.04 ≤ u ≤ 5.05 m s⁻¹
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)).
Question 18hard6 marksYear 13
A particle projected from a point on horizontal ground reaches a greatest height of 10 m and lands 40 m from its point of projection.
(a) Find the angle of projection. (4)
(b) Find the speed of projection. (2)
Mark scheme and answer
(a) \(H = \dfrac{U^2\sin^2\alpha}{2g}\), \(R = \dfrac{2U^2\sin\alpha\cos\alpha}{g}\) M1M1
\(\dfrac{H}{R} = \dfrac{\tan\alpha}{4} \Rightarrow \tan\alpha = 1\) dM1
\(\alpha = 45°\) A1
(b) \(U\sin45° = \sqrt{2\times9.8\times10}\) M1
\(U = 19.8\) m s\(^{-1}\) A1
Answer: (a) 45°; (b) 19.8 m s⁻¹
Board notes: Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)). All three boards use g = 9.8 m s⁻² unless told otherwise; answers using g = 9.8 are usually given to 2 or 3 s.f. AQA sometimes asks candidates to "show that" the trajectory equation; OCR often uses i, j vector notation for projectiles.
Question 19challenging6 marksYear 13
Particle \(A\) is projected from the origin \(O\) with velocity \((6\mathbf{i} + 20\mathbf{j})\) m s\(^{-1}\). At the same instant, particle \(B\) is released from rest at the point with position vector \((18\mathbf{i} + h\mathbf{j})\) m. Both move freely under gravity and the particles collide.
(a) Find the time at which the particles collide. (2)
(b) Find the value of \(h\). (3)
(c) Show that the collision happens while \(A\) is moving downwards. (1)
Mark scheme and answer
(a) \(6t = 18\) M1
\(t = 3\) s A1
(b) \(A\): \(y = 20(3) - 4.9(9) = 15.9\) M1
\(B\): \(y = h - 4.9(9)\) M1
\(h = 60\) A1
(c) \(v_y = 20 - 9.8(3) = -9.4 \lt 0\) B1
Answer: (a) 3 s; (b) h = 60; (c) v_y = −9.4 < 0
Board notes: Note that \(h = 20t\) independent of \(g\): both particles fall the same distance under gravity. Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)).
Question 20challenging5 marksYear 13
A particle is projected from a point on horizontal ground with speed 20 m s\(^{-1}\) at 60° above the horizontal.
(a) Find the time at which the particle is moving at right angles to its initial direction of motion. (4)
(b) Show that the particle is still in the air at this time. (1)
Mark scheme and answer
(a) \(\mathbf{v} = (20\cos60°)\mathbf{i} + (20\sin60° - gt)\mathbf{j}\) M1
Perpendicular to initial direction (60° above horizontal) means moving at 30° below the horizontal: \(\dfrac{20\sin60° - 9.8t}{20\cos60°} = -\tan30°\) M1
\(10\sqrt3 - 9.8t = -\dfrac{10}{\sqrt3}\), so \(9.8t = \dfrac{40}{\sqrt3}\) A1
\(t = 2.36\) s A1
(b) Flight time \(= \dfrac{40\sin60°}{9.8} = 3.53\) s \(\gt 2.36\) B1
Answer: (a) 2.36 s; (b) flight time 3.53 s
Board notes: A level Mathematics does not include the scalar product (it is Further Mathematics content), so the scheme uses the direction of the velocity (30° below the horizontal); a correct scalar-product method is also credited. Modelling assumptions: the object is a particle, air resistance is negligible, \(g\) is constant (9.8 m s\(^{-2}\)).
Original questions in the style of Edexcel 9MA0, independently checked before publishing. Not past-paper questions. Covers: Horizontal projection, Time of flight, range, greatest height, Modelling assumptions, Projection from a height, Clearing an obstacle, Vector form of projectile motion, Finding the speed of projection, Projection below the horizontal, Speed and direction at a point, Finding the angle of projection, Equation of the trajectory, Range formula, Time spent above a given height, Landing at a different level, Finding speed and angle, Two projectiles, Direction of motion.
A Level Math Revision