NESA Physics Projectile Motion

7 sample questions with marking guides and sample answers · Avg. score: 49.6%

Q2
2021
QCAA
Paper 1
1 mark
Q2
1 mark

Calculate the initial horizontal velocity of a projectile with an initial velocity of 38 m s−1^{-1} at an angle of 42° up from the horizontal.

A

25 m s−1^{-1}

B

28 m s−1^{-1}

C

34 m s−1^{-1}

D

40 m s−1^{-1}

Reveal Answer
A

25 m s−1^{-1}

This is the initial vertical velocity, calculated using vy=vsin⁡(θ)=38sin⁡(42∘)≈25 m s−1v_y = v \sin(\theta) = 38 \sin(42^\circ) \approx 25 \text{ m s}^{-1}.

B

28 m s−1^{-1}

Correct Answer

The horizontal component of velocity is found using the cosine of the angle: vx=vcos⁡(θ)=38cos⁡(42∘)≈28.2 m s−1v_x = v \cos(\theta) = 38 \cos(42^\circ) \approx 28.2 \text{ m s}^{-1}.

C

34 m s−1^{-1}

This value is incorrect; it does not result from applying the trigonometric component formulas to the given velocity and angle.

D

40 m s−1^{-1}

This is impossible because a component of the velocity vector (vxv_x) cannot be greater than the total magnitude of the velocity (38 m s−138 \text{ m s}^{-1}).

Q12
2021
QCAA
Paper 1
1 mark
Q12
1 mark

Calculate the maximum height reached by a projectile with an initial velocity of 15 m s−1^{-1} at an angle of 30° up from the horizontal.

A

2.87 m

B

3.83 m

C

8.61 m

D

11.5 m

Reveal Answer
A

2.87 m

Correct Answer

This is the correct maximum height derived from the formula H=v2sin⁡2(θ)2gH = \frac{v^2 \sin^2(\theta)}{2g}. Substituting the values yields (15)2(0.5)22(9.8)≈2.87 m\frac{(15)^2 (0.5)^2}{2(9.8)} \approx 2.87 \text{ m}.

B

3.83 m

This option is incorrect; it may result from a calculation error or using an incorrect fraction for the vertical component.

C

8.61 m

This answer incorrectly uses the horizontal component (cosine) instead of the vertical component (sine), calculating v2cos⁡2(θ)2g\frac{v^2 \cos^2(\theta)}{2g}.

D

11.5 m

This option ignores the launch angle and calculates the height as if the projectile were fired straight up (90∘90^\circ) using v22g\frac{v^2}{2g}.

Q20
2025
VCAA
1 mark
Q20
1 mark

Harriet and Tom were investigating how the speed, vv, of a falling object varied with the distance, ss, it had fallen.

They dropped a small steel ball, initially at rest, from the third floor of their school building. The speed of the ball was measured at six positions as it fell.

Air resistance can be ignored.

Which one of the following graphs of their data would be expected to result in a straight line through the origin?

A

vv versus ss

B

vv versus s\sqrt{s}

C

v2v^2 versus s\sqrt{s}

D

v\sqrt{v} versus ss

Reveal Answer
A

vv versus ss

The kinematic equation for an object falling from rest is v2=2gsv^2 = 2gs, meaning vv is proportional to s\sqrt{s}. A graph of vv versus ss would result in a curve, not a straight line.

B

vv versus s\sqrt{s}

Correct Answer

Using the kinematic equation v2=u2+2asv^2 = u^2 + 2as with an initial velocity of u=0u=0, we get v=2gsv = \sqrt{2g}\sqrt{s}. This shows that vv is directly proportional to s\sqrt{s}, which produces a straight line through the origin.

C

v2v^2 versus s\sqrt{s}

Based on the equation v2=2gsv^2 = 2gs, v2v^2 is directly proportional to ss, not s\sqrt{s}. Plotting v2v^2 versus s\sqrt{s} would result in a quadratic curve.

D

v\sqrt{v} versus ss

Since vv is proportional to s1/2s^{1/2}, v\sqrt{v} would be proportional to s1/4s^{1/4}. Plotting v\sqrt{v} versus ss would not produce a straight line.

Q8
2020
QCAA
Paper 1
1 mark
Q8
1 mark

An object 46 m above the ground is projected horizontally, with an initial velocity of 25 m s−1^{-1}.
Calculate the horizontal displacement of the object at the time it reaches the ground.

A

77 m

B

120 m

C

190 m

D

240 m

Reveal Answer
A

77 m

Correct Answer

Correct. The time of flight is determined by the vertical drop using t=2hg=2×469.8≈3.06t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 46}{9.8}} \approx 3.06 s. The horizontal displacement is then calculated as d=vx×t=25×3.06≈76.6d = v_x \times t = 25 \times 3.06 \approx 76.6 m, which rounds to 77 m.

B

120 m

Incorrect. This value likely results from incorrectly calculating the time of flight using the linear ratio t=hgt = \frac{h}{g} instead of the kinematic square root formula, yielding t≈4.7t \approx 4.7 s and a distance of roughly 117 m.

C

190 m

Incorrect. This large displacement implies a time of flight of roughly 7.6 s. This would be the correct answer if the projectile were on the Moon (g≈1.6g \approx 1.6 m s−2^{-2}), but it is incorrect for Earth gravity.

D

240 m

Incorrect. This answer is close to the result obtained if one forgets to take the square root when solving for time (t=2hg≈9.4t = \frac{2h}{g} \approx 9.4 s), which leads to a much larger horizontal displacement.

Q10
2020
SCSA
6 marks
Q10
6 marks

A golfer hits a ball at 37.0 m s−1\text{m s}^{-1} at 31.0° to the horizontal on a flat fairway. It travels 123 m. She wants to hit a target 135 m away, so she increases the angle at which she hits the ball, without changing the launch speed. Calculate the smallest increase of angle that allows her to reach the target. (Hint: 2sin⁡θcos⁡θ=sin⁡2θ2\sin\theta\cos\theta = \sin 2\theta)

Reveal Answer

t=135/37cos⁡Θt = 135/37 \cos\Theta
0=37sin⁡Θ−4.9(135/37cos⁡Θ)0 = 37 \sin\Theta - 4.9 (135/37 \cos\Theta)
372sin⁡Θcos⁡Θ=4.9×13537^2\sin\Theta\cos\Theta = 4.9 \times 135
sin⁡2Θ=2×4.9×135/372\sin2\Theta = 2 \times 4.9 \times 135/37^2
2Θ=75.1∘2\Theta = 75.1^\circ
Θ=37.5∘\Theta = 37.5^\circ
37.5−31=6.5∘37.5 - 31 = 6.5^\circ

Marking Criteria

Expresses t as range over horizontal velocity

Marking Bands
DescriptorMarks

expresses t as range over horizontal velocity (t = 135/37 cosΘ)

1

None of the above

0

Substitutes time into equation for vertical displacement

Marking Bands
DescriptorMarks

substitutes time into equation for vertical displacement (s = 0) and simplifies (0 = 37 sinΘ - 4.9 (135/37 cosΘ), 37²sinΘcosΘ = 4.9 × 135)

2

substitutes time into equation for vertical displacement (s = 0) (0 = 37 sinΘ - 4.9 (135/37 cosΘ))

1

None of the above

0

Solves for angle using expression given

Marking Bands
DescriptorMarks

solves for angle using expression given (sin2Θ = 2 × 4.9 × 135/37², 2Θ = 75.1°, Θ = 37.5°)

2

partially solves for angle using expression given (e.g., sin2Θ = 2 × 4.9 × 135/37²)

1

None of the above

0

Subtracts initial angle to find change of angle

Marking Bands
DescriptorMarks

subtracts initial angle to find change of angle (37.5 - 31 = 6.5°)

1

None of the above

0
Q18
2020
QCAA
Paper 1
1 mark
Q18
1 mark

Calculate the initial vertical velocity of a projectile with an initial velocity of 68 m s−1^{-1} at an angle of 51∘51^\circ up from the horizontal.

A

43 m s−1^{-1}

B

51 m s−1^{-1}

C

53 m s−1^{-1}

D

68 m s−1^{-1}

Reveal Answer
A

43 m s−1^{-1}

This is the horizontal velocity component. It is calculated using vx=vcos⁡(θ)v_x = v \cos(\theta), which gives 68cos⁡(51∘)≈43 m s−168 \cos(51^\circ) \approx 43 \text{ m s}^{-1}.

B

51 m s−1^{-1}

This value matches the angle of projection (51∘51^\circ) rather than the vertical velocity component.

C

53 m s−1^{-1}

Correct Answer

The initial vertical velocity is calculated using the sine component of the total velocity: vy=vsin⁡(θ)v_y = v \sin(\theta). Calculating 68sin⁡(51∘)68 \sin(51^\circ) yields approximately 53 m s−153 \text{ m s}^{-1}.

D

68 m s−1^{-1}

This is the magnitude of the total initial velocity vector, not just the vertical component.

Q9
2022
QCAA
Paper 2
9 marks
Q9

A person spins an object 4.3 m above the ground in a horizontal circular path of radius 0.8 m. They release the object horizontally, allowing it to travel to the ground.

Q9a
4 marks

Calculate the centripetal acceleration of the object before it is released, given it takes 5 s for the object to complete 12 revolutions. Show your working. (ms−2ms^{-2} to two significant figures)

Reveal Answer

v=2πrTv = \frac{2\pi r}{T}
=2π×0.85÷12= \frac{2\pi \times 0.8}{5 \div 12}
=12.1 m s−1= 12.1\ m\ s^{-1}

ac=v2ra_c = \frac{v^2}{r}
=12.120.8= \frac{12.1^2}{0.8}
=180 m s−2= 180\ m\ s^{-2}

Centripetal acceleration = 180 m s−2^{-2} (to two significant figures)

Marking Criteria
DescriptorMarks

Recognises the scenario relates to velocity in circular motion

1

Recognises the scenario relates to centripetal acceleration

1

Provides appropriate mathematical reasoning

1

Calculates the centripetal acceleration of the object

1
Q9b
5 marks

Calculate the total horizontal displacement for the object after it is released. Show your working. (m to two significant figures)

Reveal Answer

sy=uyt+12at2s_y = u_y t + \frac{1}{2} a t^2
4.3=0+12×9.8t24.3 = 0 + \frac{1}{2} \times 9.8 t^2
t=4.34.9t = \sqrt{\frac{4.3}{4.9}}
=0.94 s= 0.94\ s

sx=uxt+12at2s_x = u_x t + \frac{1}{2} a t^2
=12.1×0.94+12×0×0.942= 12.1 \times 0.94 + \frac{1}{2} \times 0 \times 0.94^2
=11.4 m= 11.4\ m

Horizontal displacement = 11 m (to two significant figures)

Marking Criteria
DescriptorMarks

Recognises the scenario relates to vertical component of projectile motion

1

Provides appropriate mathematical reasoning

1

Determines the time of flight

1

Recognises the scenario relates to the horizontal component of projectile motion

1

Calculates the total horizontal displacement

1

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