QCE Vault / Physics Projectile motion practice QCE Physics · Original practice questions with worked solutions
Browse all questions Projectile motion practice Quantum theory practice Original practice exam Practise resolving velocity, modelling flight and interpreting projectile diagrams. Use one consistent coordinate system and treat horizontal and vertical motion separately, linked by the same elapsed time.
Key ideas For launch speed u u u at angle θ \theta θ , use u x = u cos θ u_x=u\cos\theta u x = u cos θ and u y = u sin θ u_y=u\sin\theta u y = u sin θ when the angle is measured above the horizontal. With upward positive and negligible air resistance, x = u x t x=u_xt x = u x t and y = u y t − 1 2 g t 2 y=u_yt-\tfrac12gt^2 y = u y t − 2 1 g t 2 , measured from the launch point. At the highest point, vertical velocity is zero. Horizontal velocity remains constant under this model.
Worked example For u = 20 m s − 1 u=20\,\mathrm{m\,s^{-1}} u = 20 m s − 1 , θ = 30 ∘ \theta=30^\circ θ = 3 0 ∘ and g = 9.8 m s − 2 g=9.8\,\mathrm{m\,s^{-2}} g = 9.8 m s − 2 , u y = 10 m s − 1 u_y=10\,\mathrm{m\,s^{-1}} u y = 10 m s − 1 . The time to the highest point is 10 / 9.8 ≈ 1.02 s 10/9.8\approx1.02\,\mathrm{s} 10/9.8 ≈ 1.02 s . A common mistake The familiar range formula assumes launch and landing are at the same height. Use the vertical displacement equation when the heights differ.
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Q1 · Practice question Simple familiar 1 mark
QUESTION 1 A force of 18.0 N 18.0\,\text{N} 18.0 N acts at 35 ∘ 35^\circ 3 5 ∘ above the horizontal. What is its horizontal component? (A) 10.3 N 10.3\,\text{N} 10.3 N (B) 14.7 N 14.7\,\text{N} 14.7 N (C) 18.0 N 18.0\,\text{N} 18.0 N (D) 22.0 N 22.0\,\text{N} 22.0 N WORKED SOLUTION
Answer B 1 mark Worked solution
F x = F cos θ = 18.0 cos 35 ∘ = 14.7 N F_x=F\cos\theta=18.0\cos35^\circ=14.7\,\text{N} F x = F cos θ = 18.0 cos 3 5 ∘ = 14.7 N . Select the correct option.
[1 mark] Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabus Q57 · Practice question Complex familiar 4 marks
QUESTION 57 (4 marks) A ball is launched from level ground at 24.0 m s − 1 24.0\,\text{m s}^{-1} 24.0 m s − 1 and 35 ∘ 35^\circ 3 5 ∘ above the horizontal, as shown. Neglect air resistance. Determine (a) the time of flight and (b) the horizontal range. WORKED SOLUTION
Practice marking scheme 4 marks ANSWER t = 2.81 s t=2.81\,\text{s} t = 2.81 s ; R = 55.2 m R=55.2\,\text{m} R = 55.2 m . Worked solution
Resolve the launch velocity:
u x = 24.0 cos 35 ∘ = 19.66 m s − 1 , u y = 24.0 sin 35 ∘ = 13.77 m s − 1 . u_x=24.0\cos35^\circ=19.66\,\text{m s}^{-1},\qquad u_y=24.0\sin35^\circ=13.77\,\text{m s}^{-1}. u x = 24.0 cos 3 5 ∘ = 19.66 m s − 1 , u y = 24.0 sin 3 5 ∘ = 13.77 m s − 1 .
For a return to the launch height,
t = 2 u y g = 2 ( 13.77 ) 9.8 = 2.81 s . t=\frac{2u_y}{g}=\frac{2(13.77)}{9.8}=2.81\,\text{s}. t = g 2 u y = 9.8 2 ( 13.77 ) = 2.81 s .
Then
R = u x t = ( 19.66 ) ( 2.81 ) = 55.2 m . R=u_xt=(19.66)(2.81)=55.2\,\text{m}. R = u x t = ( 19.66 ) ( 2.81 ) = 55.2 m . Correctly resolves the initial velocity.
[1 mark] Determines the time of flight.
[1 mark] Uses horizontal motion correctly.
[1 mark] Obtains the correct range with units.
[1 mark] Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabus How many marks did you earn? Compare your working with the guide above.
Q58 · Practice question Complex familiar 5 marks
QUESTION 58 (5 marks) A stone is projected horizontally at 18.0 m s − 1 18.0\,\text{m s}^{-1} 18.0 m s − 1 from the top of a 45.0 m 45.0\,\text{m} 45.0 m cliff. Neglect air resistance. Determine the horizontal distance travelled and the magnitude and direction of its velocity immediately before impact. WORKED SOLUTION
Practice marking scheme 5 marks ANSWER x = 54.5 m x=54.5\,\text{m} x = 54.5 m ; v = 34.7 m s − 1 v=34.7\,\text{m s}^{-1} v = 34.7 m s − 1 at 58.8 ∘ 58.8^\circ 58. 8 ∘ below the horizontal. Worked solution
Vertical fall time:
45.0 = 1 2 ( 9.8 ) t 2 ⇒ t = 3.03 s . 45.0=\tfrac12(9.8)t^2\Rightarrow t=3.03\,\text{s}. 45.0 = 2 1 ( 9.8 ) t 2 ⇒ t = 3.03 s .
Horizontal distance:
x = ( 18.0 ) ( 3.03 ) = 54.5 m . x=(18.0)(3.03)=54.5\,\text{m}. x = ( 18.0 ) ( 3.03 ) = 54.5 m .
Vertical impact velocity:
v y = g t = ( 9.8 ) ( 3.03 ) = 29.7 m s − 1 downward . v_y=gt=(9.8)(3.03)=29.7\,\text{m s}^{-1}\text{ downward}. v y = g t = ( 9.8 ) ( 3.03 ) = 29.7 m s − 1 downward .
The horizontal component remains 18.0 m s − 1 18.0\,\text{m s}^{-1} 18.0 m s − 1 , so
v = 18.0 2 + 29.7 2 = 34.7 m s − 1 . v=\sqrt{18.0^2+29.7^2}=34.7\,\text{m s}^{-1}. v = 18. 0 2 + 29. 7 2 = 34.7 m s − 1 .
θ = tan − 1 ( 29.7 / 18.0 ) = 58.8 ∘ \theta=\tan^{-1}(29.7/18.0)=58.8^\circ θ = tan − 1 ( 29.7/18.0 ) = 58. 8 ∘
below the horizontal. Determines horizontal distance.
[1 mark] Determines vertical impact velocity.
[1 mark] Determines resultant speed.
[1 mark] Determines impact direction.
[1 mark] Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
View the QCAA syllabus How many marks did you earn? Compare your working with the guide above.
All projectile motion practice questions 53 original questions · Page 1 of 3
Q1 · Original practice · 1 mark A force of 18.0 N 18.0\,\text{N} 18.0 N acts at 35 ∘ 35^\circ 3 5 ∘ above the horizontal. What is its horizontal component? Projectile motion Q2 · Original practice · 1 mark A ball is launched horizontally from a platform 20.0 m 20.0\,\text{m} 20.0 m above level ground. Neglecting air resistance, how long does it take to reach the ground? Projectile motion Q3 · Original practice · 1 mark The graph shows the horizontal range of a projectile launched at constant speed on level ground as the launch angle changes. Which launch angle gives the greatest range?
Projectile motion Q11 · Original practice · 1 mark For ideal projectile motion with negligible drag, which statement is correct?
Projectile motion Q57 · Original practice · 4 marks A ball is launched from level ground at 24.0 m s − 1 24.0\,\text{m s}^{-1} 24.0 m s − 1 and 35 ∘ 35^\circ 3 5 ∘ above the horizontal, as shown. Neglect air resistance. Determine (a) the time of flight and (b) the horizontal range. Projectile motion Q58 · Original practice · 5 marks A stone is projected horizontally at 18.0 m s − 1 18.0\,\text{m s}^{-1} 18.0 m s − 1 from the top of a 45.0 m 45.0\,\text{m} 45.0 m cliff. Neglect air resistance. Determine the horizontal distance travelled and the magnitude and direction of its velocity immediately before impact. Projectile motion Q59 · Original practice · 4 marks A launcher fires identical projectiles at the same speed. The measured ranges are:
angle (°) 20 30 40 50 60 70 range (m) 31.6 40.9 46.4 46.1 41.2 31.4
Analyse the data to identify two patterns and predict the launch angle that would produce the maximum range.
Projectile motion Q102 · Original practice · 1 mark A ball is launched horizontally. Air resistance is negligible. When the ball has travelled a horizontal distance of 4.0 m 4.0\,\mathrm{m} 4.0 m , its velocity is directed 30 ∘ 30^\circ 3 0 ∘ below the horizontal. Take g = 9.8 m s − 2 g=9.8\,\mathrm{m\,s^{-2}} g = 9.8 m s − 2 . What was the ball's initial speed? Projectile motion Q106 · Original practice · 1 mark A cyclist rounds a level circular bend at constant speed. Which vector points towards the centre of the bend?
Gravity and motion Q107 · Original practice · 1 mark A satellite is moved from a circular orbit of radius r r r to a circular orbit of radius 4 r 4r 4 r around the same planet. Its orbital speed becomes Gravity and motion Q108 · Original practice · 1 mark Two projectiles are launched from level ground with the same speed at 25 ∘ 25^{\circ} 2 5 ∘ and 65 ∘ 65^{\circ} 6 5 ∘ . Neglect air resistance. Which quantity is the same for both? Gravity and motion Q125 · Original practice · 1 mark Planet B orbits the same star at 9 9 9 times Planet A’s orbital radius. The ratio T B T A \frac{T_B}{T_A} T A T B is Gravity and motion Q126 · Original practice · 3 marks A rescue capsule is launched from a 3.0 m 3.0\,\mathrm{m} 3.0 m high platform at 24.0 m s − 1 24.0\,\mathrm{m}\,\mathrm{s}^{-1} 24.0 m s − 1 and 40.0 ∘ 40.0^{\circ} 40. 0 ∘ above the horizontal. Air resistance is negligible. Gravity and motion Q131 · Original practice · 5 marks A drone moves with horizontal velocity components 6.0 m s − 1 6.0\,\mathrm{m}\,\mathrm{s}^{-1} 6.0 m s − 1 east and 8.0 m s − 1 8.0\,\mathrm{m}\,\mathrm{s}^{-1} 8.0 m s − 1 north before releasing a package from a height of 45 m 45\,\mathrm{m} 45 m . Neglect air resistance. Gravity and motion Q136 · Original practice · 8 marks A moon of radius 1.74 × 10 6 m 1.74\times10^{6}\,\mathrm{m} 1.74 × 1 0 6 m and mass 7.35 × 10 22 k g 7.35\times10^{22}\,\mathrm{kg} 7.35 × 1 0 22 kg has a research satellite in a circular orbit 100 k m 100\,\mathrm{km} 100 km above its surface. Gravity and motion Q141 · Original practice · 9 marks A ball is kicked from level ground and lands on the flat roof of a building. Air resistance is negligible.
The graph shows the speed of the ball from the instant it is kicked until the instant it lands on the roof.
Projectile motion Q145 · Original practice · 1 mark A ball is projected at 24.0 m s − 1 24.0\,\mathrm{m\,s^{-1}} 24.0 m s − 1 at 60.0 ∘ 60.0^\circ 60. 0 ∘ above horizontal. Ignore air resistance. Which is its velocity at the highest point? Gravity and motion Q146 · Original practice · 1 mark A block rests on a smooth incline at angle θ \theta θ . A force P P P is applied parallel to the incline, as shown. What is the magnitude of the normal force? Gravity and motion Q147 · Original practice · 1 mark Satellite R has mass m m m and circular orbital radius r r r . Satellite S has mass 2 m 2m 2 m and circular orbital radius 4 r 4r 4 r about the same planet. What is T S / T R T_S/T_R T S / T R ? Gravity and motion Q155 · Original practice · 8 marks A ball leaves a platform 5.00 m 5.00\,\mathrm m 5.00 m above level ground with velocity components u x = 16.0 m s − 1 u_x=16.0\,\mathrm{m\,s^{-1}} u x = 16.0 m s − 1 and u y = 12.0 m s − 1 u_y=12.0\,\mathrm{m\,s^{-1}} u y = 12.0 m s − 1 . A thin wall is 24.0 m 24.0\,\mathrm m 24.0 m horizontally from the launch point and 11.0 m 11.0\,\mathrm m 11.0 m high. Ignore drag and the ball's size. Use g = 9.8 m s − 2 g=9.8\,\mathrm{m\,s^{-2}} g = 9.8 m s − 2 . Gravity and motion Q156 · Original practice · 7 marks A camera measures a projectile's position relative to its launch point. At t = 0.500 s t=0.500\,\mathrm s t = 0.500 s , ( x , y ) = ( 6.00 , 3.275 ) m (x,y)=(6.00,3.275)\,\mathrm m ( x , y ) = ( 6.00 , 3.275 ) m . At t = 1.500 s t=1.500\,\mathrm s t = 1.500 s , ( x , y ) = ( 18.00 , 2.475 ) m (x,y)=(18.00,2.475)\,\mathrm m ( x , y ) = ( 18.00 , 2.475 ) m . Ignore air resistance and use g = 9.8 m s − 2 g=9.8\,\mathrm{m\,s^{-2}} g = 9.8 m s − 2 . Positive y y y is upward. Gravity and motion Q157 · Original practice · 6 marks A launcher fires from and lands at the same level. Its speed setting is held constant. Measured horizontal ranges have absolute uncertainty ± 0.2 m \pm0.2\,\mathrm m ± 0.2 m . Ignore drag in the theoretical model. Use g = 9.8 m s − 2 g=9.8\,\mathrm{m\,s^{-2}} g = 9.8 m s − 2 . | Launch angle | 20 ∘ 20^\circ 2 0 ∘ | 35 ∘ 35^\circ 3 5 ∘ | 55 ∘ 55^\circ 5 5 ∘ | 70 ∘ 70^\circ 7 0 ∘ |
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| Range (m) | 14.8 | 21.… Gravity and motion Q158 · Original practice · 7 marks A 5.00 k g 5.00\,\mathrm{kg} 5.00 kg crate is moving up a 30.0 ∘ 30.0^\circ 30. 0 ∘ incline. A rope pulls parallel to the incline with tension 25.0 N 25.0\,\mathrm N 25.0 N . Friction has magnitude 4.00 N 4.00\,\mathrm N 4.00 N . Its initial speed is 3.00 m s − 1 3.00\,\mathrm{m\,s^{-1}} 3.00 m s − 1 . Use g = 9.8 m s − 2 g=9.8\,\mathrm{m\,s^{-2}} g = 9.8 m s − 2 and assume the forces remain constant until it first stops. Gravity and motion Q159 · Original practice · 6 marks A 3.00 k g 3.00\,\mathrm{kg} 3.00 kg block is moving up a 25.0 ∘ 25.0^\circ 25. 0 ∘ incline. A horizontal force of 15.0 N 15.0\,\mathrm N 15.0 N pushes it to the right. Friction of 2.00 N 2.00\,\mathrm N 2.00 N acts down the slope. Use g = 9.8 m s − 2 g=9.8\,\mathrm{m\,s^{-2}} g = 9.8 m s − 2 . Gravity and motion 