References only, not question text. Open Official paper ↗ where Cambridge publishes the paper,
or see the official Cambridge past-papers page; for other papers, ask your
school. “Paper 42” means Paper 4, variant 2.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Two projectiles · Core · Paper 3 style · 5 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)Particle A is projected from a point O on horizontal ground with speed 25 m s⁻¹ at an angle α above the horizontal, where tan α = ¾. At the same instant particle B is released from rest at a point vertically above the point C on the ground, where OC = 40 m. The particles collide. Find the time at which they collide.[2]
(b)Find the height above the ground at which B was released, and the height at which the particles collide.[3]
Show answer
(a) 2 s
M1 horizontal component 25 × 0.8 = 20 m s⁻¹, 20t = 40; A1 t = 2.
(b) Released at 30 m; collision at 10 m
M1 height of A at t = 2: 15(2) − 5(4) = 10; M1 B has fallen 5(4) = 20 m; A1 h = 30 m (note tan α = 30/40: A was aimed directly at B's starting point).
Q2Two projectiles · Core · Paper 3 style · 7 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)Points A and B are on horizontal ground, 60 m apart. Particle P is projected from A towards B with speed 26 m s⁻¹ at an angle θ above the horizontal, where tan θ = 12/5. At the same instant particle Q is projected from B towards A with initial velocity components 20 m s⁻¹ horizontally and 24 m s⁻¹ vertically upwards. Show that P and Q are always at the same height while both are in the air.[2]
(b)Find the time at which the particles collide and the height of the collision.[3]
(c)Find the speed and angle of projection of Q.[2]
Show answer
(a) Shown
M1 P's vertical component 26 × 12/13 = 24 m s⁻¹; A1 both have height 24t − 5t².
(b) t = 2 s; 28 m
M1 10t + 20t = 60; A1 t = 2; A1 24(2) − 5(4) = 28 m.
(c) 31.2 m s⁻¹ at 50.2° above the horizontal
M1 √(20² + 24²); A1 tan⁻¹(24/20).
Q3Two projectiles · Core · Paper 3 style · 3 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)Two stones are thrown horizontally, in the same direction, from the same point at the top of a tower 20 m high. One has speed 8 m s⁻¹ and the other 14 m s⁻¹. Find the time each stone takes to reach the ground.[1]
(b)Find the distance between the points where the stones land.[2]
Show answer
(a) 2 s
B1 20 = 5t², t = 2 for both.
(b) 12 m
M1 14 × 2 − 8 × 2; A1 12 m.
Q4Two projectiles · Standard · Paper 3 style · 7 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A stone A is thrown horizontally with speed 12 m s⁻¹ from the top of a vertical cliff 45 m above the sea. Find the horizontal distance from the foot of the cliff to the point where A hits the sea.[2]
(b)One second after A is thrown, a second stone B is thrown from the same point with speed 20 m s⁻¹ at 30° above the horizontal, in the same vertical plane and the same horizontal direction as A. Find the distance between the points where the two stones hit the sea, and the time between their impacts.[5]
Show answer
(a) 36 m
M1 45 = 5t², t = 3; A1 12 × 3 = 36.
(b) 36.1 m; 2.16 s
M1 −45 = 10t − 5t² for B; M1 t² − 2t − 9 = 0; A1 t = 1 + √10 = 4.162 s after B is thrown; M1 horizontal distance 10√3 × 4.162 = 72.09 m; A1 separation 36.1 m; B lands 5.162 s after A was thrown, i.e. 2.16 s after A lands.
Q5Two projectiles · Standard · Paper 3 style · 6 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)Particle P is projected vertically upwards from a point A on horizontal ground with speed 30 m s⁻¹. Find the time taken for P to reach its greatest height, and that height.[2]
(b)At the instant P is projected, a particle Q is projected from a point B on the ground, 40 m from A, with speed u m s⁻¹ at an angle θ above the horizontal. Q hits P when P is at its greatest height. Find u and θ.[4]
Show answer
(a) 3 s; 45 m
B1 t = 30/10; B1 30²/20 = 45.
(b) u = 32.8, θ = 66.0°
M1 horizontal: 3u cos θ = 40; M1 vertical: 3u sin θ − 45 = 45, so u sin θ = 30; A1 u = √(30² + (40/3)²) = 32.8; A1 tan θ = 2.25, θ = 66.0°.
Q6Two projectiles · Standard · Paper 3 style · 6 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)Particle A is projected horizontally with speed 15 m s⁻¹ from the top of a vertical tower of height 40 m. At the same instant particle B is projected from the foot of the tower with speed V m s⁻¹ at an angle θ above the horizontal, in the same vertical plane and direction as A. The particles collide. Show that V cos θ = 15.[2]
(b)The collision occurs 2 s after projection. Find V, θ and the height above the ground of the collision.[4]
Show answer
(a) Shown
M1 both start on the same vertical line, so equal horizontal displacements at all times; A1 V cos θ t = 15t (AG).
(b) V = 25, θ = 53.1°, height 20 m
M1 vertical separation: 40 − 5t² = V sin θ t − 5t², so V sin θ × 2 = 40; A1 V sin θ = 20; A1 V = 25, θ = tan⁻¹(4/3) = 53.1°; A1 40 − 5(4) = 20 m (A has not yet landed, since it lands at √8 ≈ 2.83 s).
Q7Two projectiles · Challenge · Paper 3 style · 7 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)Two golf balls are struck from the same point O on horizontal ground, each with speed 30 m s⁻¹, at angles α and β to the horizontal, where α < β. Both land at the same point, 72 m from O. Show that sin 2α = 0.8 and find α and β.[4]
(b)The ball at angle β is struck first. Find how long after it the other ball must be struck so that both land at the same instant.[3]
Show answer
(a) α = 26.6°, β = 63.4°
M1 range u² sin 2θ/g = 72; A1 sin 2θ = 720/900 = 0.8 (AG); A1 α = ½ sin⁻¹ 0.8; A1 β = 90° − α.
(b) 2.68 s
M1 time of flight 2u sin θ/g = 6 sin θ; M1 sin α = 1/√5, sin β = 2/√5, so times 2.68 s and 5.37 s; A1 difference 6/√5 = 2.68 s.
Q8Two projectiles · Challenge · Paper 3 style · 8 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)Particle A is projected from a point O on horizontal ground with velocity components 10 m s⁻¹ horizontally and 25 m s⁻¹ vertically. One second later particle B is projected from O, in the same vertical plane, with velocity components 15 m s⁻¹ horizontally and 25 m s⁻¹ vertically. Show that the particles collide, and find the time after A is projected and the position of the collision.[5]
(b)Find the angle between the directions of motion of A and B at the moment of collision.[3]
Show answer
(a) t = 3 s after A is projected; at 30 m horizontally from O and 30 m above the ground
M1 horizontal: 10t = 15(t − 1); A1 t = 3; M1 A's height 25(3) − 5(9) = 30; M1 B's height 25(2) − 5(4) = 30; A1 equal, so they collide at (30, 30).
Q9Two projectiles · Challenge · Paper 3 style · 7 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)Two balls are kicked at the same instant from the same point O on horizontal ground. Ball P has initial velocity components 18 m s⁻¹ horizontally and 24 m s⁻¹ vertically; ball Q, in the same vertical plane and direction, has components 24 m s⁻¹ horizontally and 7 m s⁻¹ vertically. Show that, while both are in the air, the distance between them increases at a constant rate, and find this rate.[4]
(b)Find which ball lands first, and the distance between the balls at that instant.[3]
Show answer
(a) √325 = 18.0 m s⁻¹
M1 positions (18t, 24t − 5t²) and (24t, 7t − 5t²); M1 difference (6t, −17t) has no t² term; A1 distance t√(6² + 17²); A1 rate √325.
(b) Q, after 1.4 s; 25.2 m
M1 flight times 2 × 24/10 = 4.8 s and 2 × 7/10 = 1.4 s; A1 Q lands first; A1 1.4√325 = 25.2 m.
Q10Equation of trajectory · Core · Paper 3 style · 5 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A particle is projected from a point O on horizontal ground with speed 20 m s⁻¹ at 45° above the horizontal. Taking O as origin, x horizontal and y vertically upwards (in metres), show that the equation of its trajectory is y = x − x²/40.[3]
(b)Find the horizontal distances from O at which the particle is 7.5 m above the ground.[2]
Show answer
(a) Shown
M1 x = 20 cos 45° t, y = 20 sin 45° t − 5t²; M1 eliminate t = x/(10√2); A1 y = x − 5x²/200 (AG).
(b) 10 m and 30 m
M1 x² − 40x + 300 = 0; A1 x = 10 or 30.
Q11Equation of trajectory · Core · Paper 3 style · 5 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A ball is kicked from a point O on horizontal ground. Its trajectory has equation y = 0.75x − 0.02x², where x and y are the horizontal and vertical distances in metres from O. Find the angle of projection.[1]
(b)Find the speed of projection.[3]
(c)Find the horizontal distance from O to where the ball lands.[1]
Show answer
(a) 36.9°
B1 tan θ = 0.75.
(b) 19.8 m s⁻¹
M1 g/(2u² cos² θ) = 0.02; M1 cos θ = 0.8 so u² = 10/(2 × 0.64 × 0.02) = 390.625; A1 u = 19.8.
(c) 37.5 m
B1 0.75/0.02.
Q12Equation of trajectory · Core · Paper 3 style · 5 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A ball is thrown from a point O on horizontal ground at 45° above the horizontal and hits a vertical wall at a point 3 m above the ground. The wall is 12 m from O. Using the equation of the trajectory, find the speed of projection.[3]
(b)Determine whether the ball is still rising when it hits the wall.[2]
Show answer
(a) 12.6 m s⁻¹
M1 y = x − gx²/u² for θ = 45°; M1 3 = 12 − 1440/u²; A1 u² = 160, u = 12.6.
(b) No: the highest point is at x = u²/(2g) = 8 m < 12 m, so the ball is descending
M1 dy/dx = 1 − 20x/u² at x = 12 is −0.5 (or compare with x of top); A1 descending.
Q13Equation of trajectory · Standard · Paper 3 style · 7 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A ball is projected from a point O on horizontal ground with speed 25 m s⁻¹ at an angle θ above the horizontal. Show that, when the ball has travelled a horizontal distance x m, its height y m is given by y = x tan θ − x²(1 + tan² θ)/125.[3]
(b)A vertical wall of height 4 m stands 30 m from O, perpendicular to the plane of motion. Find the range of values of θ for which the ball passes over the wall.[4]
Show answer
(a) Shown
M1 x = 25 cos θ t, y = 25 sin θ t − 5t²; M1 substitute t; A1 sec² θ = 1 + tan² θ and 5/625 = 1/125 (AG).
Q14Equation of trajectory · Standard · Paper 3 style · 5 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A stone is thrown from a point O with speed 30 m s⁻¹ at an angle θ above the horizontal, and passes through a target that is 40 m horizontally from O and 10 m above O. Show that 8 tan² θ − 36 tan θ + 17 = 0.[3]
(b)Find the two possible angles of projection.[2]
Show answer
(a) Shown
M1 trajectory y = x tan θ − gx²(1 + tan² θ)/(2u²); M1 10 = 40T − 16000(1 + T²)/1800; A1 multiply by 9/10 and rearrange (AG).
(b) 75.8° and 28.2°
M1 tan θ = 3.964 or 0.5361; A1 both angles.
Q15Equation of trajectory · Standard · Paper 3 style · 6 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A stone is thrown from a point O at the top of a cliff, 50 m above the sea, with speed 20 m s⁻¹ at 30° above the horizontal. Taking O as origin with y measured vertically upwards, find the equation of the trajectory in the form y = ax − bx².[3]
(b)Find the horizontal distance from O at which the stone hits the sea.[3]
Show answer
(a) y = x/√3 − x²/60
M1 x = 10√3 t, y = 10t − 5t²; M1 t = x/(10√3); A1 y = x/√3 − x²/60.
Q16Equation of trajectory · Standard · Paper 3 style · 5 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A basketball player releases the ball 2.1 m above the floor with speed 9 m s⁻¹ at an angle θ above the horizontal. The ball passes through the centre of the hoop, which is 3.05 m above the floor and 6 m horizontally from the point of release. Treating the ball as a particle, show that 20 tan² θ − 54 tan θ + 28.55 = 0.[3]
(b)Find the two possible values of θ.[2]
Show answer
(a) Shown
M1 relative height 0.95 m; M1 0.95 = 6 tan θ − 10(36)(1 + tan² θ)/(2 × 81); A1 multiply by 9 and rearrange: 20T² − 54T + 28.55 = 0 (AG).
(b) 63.2° or 35.8°
M1 tan θ = 1.978 or 0.7215; A1 both.
Q17Equation of trajectory · Challenge · Paper 3 style · 7 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A particle is projected from O with speed u m s⁻¹ at an angle θ above the horizontal. It passes through the point P, 30 m horizontally from O and 20 m above O. Show that 4500 tan² θ − 30u² tan θ + (4500 + 20u²) = 0.[3]
(b)By considering the discriminant of this quadratic in tan θ, find the least possible value of u, and the corresponding angle of projection.[4]
Show answer
(a) Shown
M1 20 = 30 tan θ − 10(900)(1 + tan² θ)/(2u²); M1 multiply by u²; A1 rearrange (AG).
Q18Equation of trajectory · Challenge · Paper 3 style · 7 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A particle is projected from a point O on horizontal ground and moves freely under gravity. Its trajectory passes through the points (10, 8) and (30, 12), where coordinates are in metres with O as origin and y vertically upwards. Using the trajectory equation in the form y = x tan θ − kx², where k = g/(2u² cos² θ), find tan θ and k.[4]
(b)Find the speed of projection and the range on the horizontal ground.[3]
Show answer
(a) tan θ = 1, k = 0.02
M1 8 = 10 tan θ − 100k and 12 = 30 tan θ − 900k; M1 eliminate tan θ: 3 × first − second gives 24 − 12 = −300k + 900k = 600k; A1 k = 0.02; A1 tan θ = 1.
(b) 22.4 m s⁻¹; 50 m
M1 10/(2u² × 0.5) = 0.02; A1 u = √500 = 22.4; B1 range tan θ/k = 50 m.
Q19Direction & speed during flight · Core · Paper 3 style · 5 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A particle is projected from a point on horizontal ground with speed 25 m s⁻¹ at an angle α above the horizontal, where tan α = 4/3. Find the speed of the particle 3 s after projection.[3]
(b)Find the direction of motion at this instant.[2]
Q20Direction & speed during flight · Core · Paper 3 style · 4 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A ball is projected from a point O on horizontal ground with speed 20 m s⁻¹ at an angle θ above the horizontal, where sin θ = 0.8. Find the times at which the ball is moving at 45° to the horizontal.[3]
(b)Show that both times occur while the ball is in the air.[1]
Show answer
(a) 0.4 s (rising) and 2.8 s (falling)
M1 horizontal component 12, vertical 16 − 10t; M1 16 − 10t = ±12; A1 t = 0.4 and 2.8.
(b) Shown
B1 time of flight 3.2 s > 2.8 s.
Q21Direction & speed during flight · Core · Paper 3 style · 4 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A particle is projected from a point on horizontal ground with speed 20 m s⁻¹. The least speed of the particle during its flight is 12 m s⁻¹. Find the angle of projection.[2]
(b)Find the greatest height reached.[2]
Show answer
(a) 53.1°
M1 least speed = u cos α (at the highest point); A1 cos α = 0.6.
(b) 12.8 m
M1 ½m(20² − 12²) = mgH; A1 H = 256/20 = 12.8.
Q22Direction & speed during flight · Standard · Paper 3 style · 7 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A particle is projected from a point O on horizontal ground with speed 30 m s⁻¹ at 60° above the horizontal. Use conservation of energy to find the speed of the particle when it is 25 m above the ground.[2]
(b)Find the angle that the direction of motion makes with the horizontal at these instants.[2]
(c)Find the two times at which the particle is 25 m above the ground.[3]
Show answer
(a) 20 m s⁻¹
M1 v² = 30² − 2(10)(25); A1 v = 20.
(b) 41.4° (above the horizontal when rising, below when falling)
M1 horizontal component 15 throughout, cos φ = 15/20; A1.
(c) 1.28 s and 3.92 s
M1 25 = 15√3 t − 5t²; M1 t² − 3√3t + 5 = 0; A1 t = 1.28, 3.92.
Q23Direction & speed during flight · Standard · Paper 3 style · 6 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A stone is thrown from the top of a cliff 40 m above the sea with speed 15 m s⁻¹ at an angle θ above the horizontal, where tan θ = ¾. Find the time taken for the stone to reach the sea.[3]
(b)Find the speed and direction of motion of the stone as it hits the sea.[3]
Show answer
(a) 3.87 s
M1 components 12 and 9; M1 −40 = 9t − 5t²; A1 t = (9 + √881)/10.
Q24Direction & speed during flight · Challenge · Paper 3 style · 7 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A particle is projected from a point O on horizontal ground with speed u at an angle α above the horizontal. Show that its velocity is perpendicular to its initial velocity at time t = u/(g sin α).[3]
(b)Show that this happens before the particle lands only if α > 45°.[2]
(c)For u = 20 m s⁻¹ and α = 60°, find the time and the speed of the particle at this instant.[2]
Show answer
(a) Shown
M1 velocity (u cos α, u sin α − gt); M1 scalar product with (u cos α, u sin α) = 0; A1 u² − ugt sin α = 0 (AG).
(b) Shown
M1 u/(g sin α) < 2u sin α/g; A1 sin² α > ½, so α > 45° (AG).
Q25Direction & speed during flight · Challenge · Paper 3 style · 7 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A projectile passes through a point A moving at 30° above the horizontal. It passes through a point B 1.5 s later, moving at 45° below the horizontal. Find the horizontal component of its velocity and its speed at A.[4]
(b)Find the distance AB.[3]
Show answer
(a) Horizontal component 9.51 m s⁻¹; speed at A 11.0 m s⁻¹
M1 vertical components h tan 30° at A and −h at B; M1 h tan 30° + h = 10 × 1.5; A1 h = 9.51; A1 speed at A = h/cos 30° = 11.0.
(b) 14.6 m
M1 horizontal 1.5h = 14.26; M1 vertical displacement = average vertical velocity × 1.5 = −3.014 m; A1 AB = 14.6 m.
Q26Range, height & time of flight · Core · Paper 3 style · 5 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A particle is projected from a point on horizontal ground with speed u at an angle α above the horizontal. Show that its range on the ground is u² sin 2α/g.[3]
(b)A javelin is thrown with speed 35 m s⁻¹ from ground level. Treating it as a particle, find the greatest possible range.[2]
Show answer
(a) Shown
M1 time of flight 2u sin α/g; M1 range = u cos α × 2u sin α/g; A1 2 sin α cos α = sin 2α (AG).
(b) 122.5 m
M1 α = 45°; A1 35²/10 = 122.5.
Q27Range, height & time of flight · Core · Paper 3 style · 5 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A ball is projected from a point on horizontal ground with speed 40 m s⁻¹ at 30° above the horizontal. Find the time of flight.[2]
(b)Find the greatest height reached and the range.[3]
Show answer
(a) 4 s
M1 0 = 20t − 5t²; A1 t = 4.
(b) 20 m; range 80√3 m = 139 m (3 s.f.)
B1 20²/20 = 20 m; M1 20√3 × 4; A1 80√3 = 138.56… = 139 m.
Q28Range, height & time of flight · Standard · Paper 3 style · 5 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A shot-putter's practice ball is projected from ground level with speed 25 m s⁻¹ and lands 60 m away on horizontal ground. Find the two possible angles of projection.[3]
(b)Find the time of flight for each angle.[2]
Show answer
(a) 36.9° or 53.1°
M1 625 sin 2α/10 = 60; A1 sin 2α = 0.96; A1 2α = 73.74° or 106.26°, so α = 36.9° or 53.1°.
(b) 3 s and 4 s
M1 T = 2(25 sin α)/10 = 5 sin α with sin α = 0.6 or 0.8; A1 3 s and 4 s.
Q29Range, height & time of flight · Standard · Paper 3 style · 5 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A cricket ball is hit from a point 1.8 m above horizontal ground with speed 28 m s⁻¹ at 35° above the horizontal. Find the greatest height of the ball above the ground.[2]
(b)Find the horizontal distance travelled before the ball hits the ground.[3]
Show answer
(a) 14.7 m
M1 (28 sin 35°)²/20; A1 add 1.8.
(b) 76.2 m
M1 −1.8 = 28 sin 35° t − 5t²; A1 t = 3.32; A1 28 cos 35° × t = 76.2.
Q30Range, height & time of flight · Challenge · Paper 3 style · 7 marks
Take g = 10 m s⁻² and ignore air resistance. Give non-exact answers correct to 3 significant figures and angles correct to 1 decimal place.
(a)A particle is projected from a point on horizontal ground with speed u at an angle α above the horizontal. Its greatest height is H and its range is R. Show that R = 4H cot α.[3]
(b)Given that the greatest height equals the range, find tan α. For u = 20 m s⁻¹ find this height and the time of flight.[4]
Show answer
(a) Shown
M1 H = u² sin² α/(2g), R = 2u² sin α cos α/g; M1 R/H = 4 cos α/sin α; A1 (AG).
(b) tan α = 4; H = R = 18.8 m; time 3.88 s
B1 tan α = 4; M1 sin² α = 16/17; A1 H = 400(16/17)/20 = 18.8; A1 T = 2(20)(4/√17)/10 = 3.88 s.
Similar questions
Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.
References only: find each question in its paper (official link where available, otherwise through your school).
