Projectiles - Edexcel A-Level Maths (Paper 3) Exam Questions

Practise 19 exam-style questions on Projectiles for Edexcel A-Level Maths (Paper 3). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

19 questions
3–10 marks each
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19 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

5 marks

A hockey player lifts the ball from the point \(A\) on a horizontal pitch. The ball leaves the ground at \(A\) with speed \(17\ \text{m s}^{-1}\), directed upwards at an angle \(\alpha\) to the horizontal, where \(\tan\alpha = \dfrac{15}{8}\). It moves freely under gravity and first lands at the point \(B\) on the pitch. Figure 1 shows the path of the ball. The ball is modelled as a particle.

Find the time that the ball takes to travel from \(A\) to \(B\), and the distance \(AB\).

Use \(g = 9.8\ \text{m s}^{-2}\) and give your answers to 2 or 3 significant figures.

Figure 1: a horizontal pitch. A ball leaves point A on the ground with speed 17 m s⁻¹ at an angle α above the horizontal. A dashed curved path rises to a highest point and comes down to land at point B on the ground, to the right of A.

Model answer

\(\sin\alpha = \dfrac{15}{17}\), \(\cos\alpha = \dfrac{8}{17}\), so the initial velocity has components 8 (horizontal) and 15 (vertical).
Vertical, \(A\) to \(B\): \(0 = 15t - 4.9t^2\), \(t ≠ 0\)
\(t = \dfrac{15}{4.9} = 3.06\) s (3 s.f.)
Horizontal: \(AB = 8 \times 3.0612... = 24.5\) m (3 s.f.)

Explanation

Use the 8-15-17 triangle to resolve the velocity into horizontal and vertical components of 8 and 15. The time of flight comes from zero vertical displacement, and the horizontal distance is the constant horizontal speed times this time. Common errors are not resolving the 17 and giving more than 3 significant figures.

Mark scheme

1 mark
M1 Uses vertical motion from \(A\) to \(B\) with zero vertical displacement and the resolved component \(17\sin\alpha = 15\): \(0 = 15t - 4.9t^2\); or correct final answers of 3.1 s (3.06 s) and 24 m (24.5 m) with no working
Correct answers of 3.1 s (or 3.06 s) and 24 m (or 24.5 m) with no working score 2 marks (the A1 for each answer). Condone sin/cos confusion and sign errors for this M mark, but the speed must be resolved: using the full speed as a component is M0. \(\sin\alpha = \dfrac{15}{17}\), \(\cos\alpha = \dfrac{8}{17}\). Accept equivalent forms.
1 mark
A1 Correct equation: \(0 = 15t - 4.9t^2\) oe
Accept equivalent forms.
1 mark
A1 \(t = \dfrac{30}{9.8} = 3.1\) s or 3.06 s
Depends on the M mark. awrt 3.06, i.e. in [3.055, 3.065), or 3.1. Answers from g = 9.8 should be given to 2 or 3 s.f.; a more accurate final answer loses this A mark (over-accuracy is penalised once per question). Use of g = 9.81 is penalised once per question.
1 mark
M1 Uses horizontal motion: \(AB = (17\cos\alpha) \times \text{their } t = 8 \times \text{their } t\)
The horizontal component must be resolved.
1 mark
A1 \(AB = 24\) m or 24.5 m
awrt 24.5, i.e. in [24.45, 24.55), or 24. Answers from g = 9.8 should be given to 2 or 3 s.f.; a more accurate final answer loses this A mark (over-accuracy is penalised once per question). Use of g = 9.81 is penalised once per question.

Question 2

7 marks

A ball is thrown from a point \(A\), which is \(h\) metres vertically above a point \(O\) on horizontal ground. The ball is thrown with speed \(U\ \text{m s}^{-1}\), directed upwards at an angle \(\alpha\) to the horizontal, and moves freely under gravity in a vertical plane through \(O\). At time \(t\) seconds after it is thrown, the ball is \(x\) metres horizontally from \(O\) and \(y\) metres above the ground.

Show that, while the ball is in the air,
\[y = h + x\tan\alpha - \dfrac{gx^2(1 + \tan^2\alpha)}{2U^2}\]

Model answer

Horizontal: \(x = (U\cos\alpha)t\), so \(t = \dfrac{x}{U\cos\alpha}\)
Vertical (up positive, measured from the ground): \(y = h + (U\sin\alpha)t - \frac{1}{2}gt^2\)
Substituting for \(t\): \(y = h + U\sin\alpha \times \dfrac{x}{U\cos\alpha} - \dfrac{g}{2} \times \dfrac{x^2}{U^2\cos^2\alpha}\)
\(y = h + x\tan\alpha - \dfrac{gx^2}{2U^2\cos^2\alpha}\)
Since \(\dfrac{1}{\cos^2\alpha} = \sec^2\alpha = 1 + \tan^2\alpha\):
\(y = h + x\tan\alpha - \dfrac{gx^2(1 + \tan^2\alpha)}{2U^2}\)

Explanation

Write the horizontal and vertical displacements in terms of \(t\), then eliminate \(t\) using the horizontal equation. The identity \(\sec^2\alpha \equiv 1 + \tan^2\alpha\) turns the \(\cos^2\alpha\) term into the printed form. Common errors are forgetting the starting height \(h\) and quoting the path equation instead of deriving it.

Mark scheme

1 mark
B1 Horizontal motion: \(x = (U\cos\alpha)t\). An answer with no working scores 0 marks
An answer with no working scores 0 marks: this is a show that. Quoting the equation of the path scores no marks.
1 mark
M1 Vertical motion: \(y = h + (U\sin\alpha)t - \frac{1}{2}gt^2\) (correct number of terms, velocity resolved)
Condone sign errors and sin/cos confusion for this mark. M0 if \(h\) is missing and \(y\) is measured from the ground.
1 mark
A1 Correct vertical equation
1 mark
M1 (dM1) Eliminates \(t\) using \(t = \dfrac{x}{U\cos\alpha}\) from their horizontal equation
Depends on the M mark for the vertical equation (not on the A mark just before it).
1 mark
A1 \(y = h + x\tan\alpha - \dfrac{gx^2}{2U^2\cos^2\alpha}\)
\(\dfrac{\sin\alpha}{\cos\alpha}\) replaced by \(\tan\alpha\).
1 mark
M1 Uses \(\dfrac{1}{\cos^2\alpha} = \sec^2\alpha = 1 + \tan^2\alpha\)
Independent of the previous A mark.
1 mark
A1* \(y = h + x\tan\alpha - \dfrac{gx^2(1 + \tan^2\alpha)}{2U^2}\)
cso, with no errors in the working.

Question 3

7 marks

In this question \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors, horizontal and vertically upwards respectively.

A particle \(P\) is projected from a fixed point \(O\) on horizontal ground with velocity \((7.5\mathbf{i} + 10\mathbf{j})\ \text{m s}^{-1}\) and moves freely under gravity.

There are two points on the path of \(P\) at which the speed of \(P\) is \(8.5\ \text{m s}^{-1}\). Find the position vectors of these two points relative to \(O\), and the distance between them.

Use \(g = 9.8\ \text{m s}^{-2}\) and give your answers to 2 or 3 significant figures.

Model answer

\(\mathbf{a} = -9.8\mathbf{j}\), so \(\mathbf{v} = 7.5\mathbf{i} + (10 - 9.8t)\mathbf{j}\)
Speed 8.5: \(7.5^2 + (10 - 9.8t)^2 = 8.5^2\), so \((10 - 9.8t)^2 = 72.25 - 56.25 = 16\)
\(10 - 9.8t = 4\) or \(-4\), so \(t = \dfrac{6}{9.8} = 0.612...\) or \(t = \dfrac{14}{9.8} = 1.428...\)
\(\mathbf{r} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2 = 7.5t\,\mathbf{i} + (10t - 4.9t^2)\mathbf{j}\)
\(t = 0.612...\): \(\mathbf{r} = (4.59\mathbf{i} + 4.29\mathbf{j})\) m (3 s.f.)
\(t = 1.428...\): \(\mathbf{r} = (10.7\mathbf{i} + 4.29\mathbf{j})\) m (3 s.f.)
Both points are \(\dfrac{30}{7} = 4.29\) m above the ground, so the distance between them is \(10.714... - 4.591... = 6.12\) m (3 s.f.).

Explanation

The horizontal component of the velocity stays \(7.5\ \text{m s}^{-1}\), so a speed of \(8.5\ \text{m s}^{-1}\) needs a vertical component of \(4\ \text{m s}^{-1}\) upwards or downwards: once while rising and once while falling. By the symmetry of the path the two points are at the same height, so the distance between them is the difference of their horizontal displacements. A common error is to find only the point on the way up.

Mark scheme

1 mark
M1 Velocity at time \(t\): \(\mathbf{v} = 7.5\mathbf{i} + (10 - 9.8t)\mathbf{j}\). Correct final answers of \((4.59\mathbf{i} + 4.29\mathbf{j})\) m, \((10.7\mathbf{i} + 4.29\mathbf{j})\) m and 6.12 m with no working score 2 marks (the A1 for the position vectors and the final A1)
Correct answers of \((4.59\mathbf{i} + 4.29\mathbf{j})\) m and \((10.7\mathbf{i} + 4.29\mathbf{j})\) m and a distance of 6.12 m (or 6.1 m) with no working score 2 marks: this is a multi-step problem and the method must be seen for the other marks. Column vectors are accepted throughout. Condone a sign error in the vertical component for this M mark.
1 mark
M1 (dM1) Uses the speed: \(7.5^2 + (10 - 9.8t)^2 = 8.5^2\), so \((10 - 9.8t)^2 = 16\) (or the vertical component is \(\pm 4\))
M0 for adding the components without squaring, or for \(8.5 = 7.5 + \ldots\).
1 mark
A1 \(t = \dfrac{6}{9.8} = 0.612\) and \(t = \dfrac{14}{9.8} = 1.43\)
Both values needed: awrt 0.612, i.e. [0.6115, 0.6125), and awrt 1.43, i.e. [1.425, 1.435), or the exact fractions \(\dfrac{6}{9.8}\) and \(\dfrac{14}{9.8}\).
1 mark
M1 Uses \(\mathbf{r} = \mathbf{u}t + \tfrac{1}{2}\mathbf{a}t^2\) (or the horizontal and vertical displacements) with their values of \(t\)
Depends on the first M mark. \(\mathbf{r} = 7.5t\,\mathbf{i} + (10t - 4.9t^2)\mathbf{j}\).
1 mark
A1 \((4.59\mathbf{i} + 4.29\mathbf{j})\) m and \((10.7\mathbf{i} + 4.29\mathbf{j})\) m
Both position vectors are needed for this mark: awrt 4.59, i.e. [4.585, 4.595), and awrt 10.7, i.e. [10.65, 10.75), for the \(\mathbf{i}\)-components and awrt 4.29, i.e. [4.285, 4.295), for the \(\mathbf{j}\)-components (the exact height is \(\dfrac{30}{7}\)); 2 s.f. values are also accepted. Answers from \(g = 9.8\) should be given to 2 or 3 s.f.; use of \(g = 9.81\) is penalised once in this question.
1 mark
M1 Finds the distance between their two points (the difference of the \(\mathbf{i}\)-components if the \(\mathbf{j}\)-components are equal, or Pythagoras)
Their two points from the same method.
1 mark
A1 6.12 m or 6.1 m
awrt 6.12, i.e. in [6.115, 6.125), or 6.1. A more accurate final answer is penalised once in this question; use of \(g = 9.81\) is penalised once in this question.

Question 4

9 marks

A ball is thrown from a point \(A\) on the flat roof of a building. The point \(A\) is 40 m vertically above the point \(B\) on horizontal ground. The ball is thrown with speed \(U\ \text{m s}^{-1}\), directed upwards at an angle \(\alpha\) to the horizontal, where \(\tan\alpha = \dfrac{7}{24}\), and it moves freely under gravity in a vertical plane through \(B\). The ball hits the ground at the point \(C\). Immediately before the ball hits the ground, its direction of motion makes an angle of \(60^\circ\) with the horizontal, as shown in Figure 1. The ball is modelled as a particle.

Find the value of \(U\) and the distance \(BC\).

Use \(g = 9.8\ \text{m s}^{-2}\) and give your answers to 2 or 3 significant figures.

Figure 1: a building with a flat roof stands on horizontal ground. Point A is on the edge of the roof, vertically above point B at the foot of the building; a dimension line shows AB = 40 m. A ball is thrown from A with speed U m s⁻¹ at an angle α above the horizontal. A dashed curved path rises slightly and then falls to the ground at point C, to the right of B. At C, the direction of motion makes an angle of 60° with the ground.

Model answer

\(\sin\alpha = \dfrac{7}{25}\), \(\cos\alpha = \dfrac{24}{25}\): the initial velocity has components \(\dfrac{24U}{25}\) (horizontal) and \(\dfrac{7U}{25}\) (vertically upwards).
Vertical, \(A\) to \(C\) (40 m downwards), with \(v_y\) the downward vertical component at \(C\):
\(v_y^2 = \left(\dfrac{7U}{25}\right)^2 + 2 \times 9.8 \times 40 = \dfrac{49U^2}{625} + 784\)
The horizontal component is still \(\dfrac{24U}{25}\), so at \(C\): \(\tan 60^\circ = \dfrac{v_y}{24U/25}\), so \(v_y^2 = 3 \times \dfrac{576U^2}{625}\)
\(\dfrac{1728U^2}{625} = \dfrac{49U^2}{625} + 784\), so \(\dfrac{1679U^2}{625} = 784\)
\(U^2 = 291.8\), so \(U = 17.1\) (3 s.f.)
Components: horizontal \(\dfrac{24U}{25} = 16.40\ \text{m s}^{-1}\); vertical \(\dfrac{7U}{25} = 4.783\ \text{m s}^{-1}\) upwards at \(A\) and \(v_y = 16.40\tan 60^\circ = 28.41\ \text{m s}^{-1}\) downwards at \(C\)
Time: \(-28.41 = 4.783 - 9.8t\), so \(t = 3.387\) s
\(BC = 16.40 \times 3.387 = 55.5\) m (3 s.f.)

Explanation

The horizontal component of velocity, \(\dfrac{24U}{25}\), does not change, and the vertical component at \(C\) follows from \(v^2 = u^2 + 2as\) over the 40 m drop; the 60° direction makes the vertical component \(\tan 60^\circ\) times the horizontal one, which gives an equation in \(U\). The time of flight then comes from \(v = u + at\) vertically, and \(BC\) is the horizontal component times this time. Common errors are not resolving \(U\), using the angle of the line \(AC\) instead of the direction of motion, and rounding \(U\) too early.

Mark scheme

1 mark
M1 Uses vertical motion from \(A\) to \(C\) to find the vertical component of the velocity at \(C\) in terms of \(U\): \(v^2 = (U\sin\alpha)^2 + 2 \times 9.8 \times 40\) (or \(v = u + at\) with the time from \(-40 = (U\sin\alpha)t - 4.9t^2\)); or correct final answers of \(U = 17\) (17.1) and \(BC = 56\) m (55.5 m) with no working
Correct answers of \(U = 17\) (or 17.1) and \(BC = 56\) m (or 55.5 m) with no working score 2 marks (the A1 for each answer). Condone sin/cos confusion and sign errors for this M mark, but the speed must be resolved: using the full speed as a component is M0. \(\sin\alpha = \dfrac{7}{25}\), \(\cos\alpha = \dfrac{24}{25}\). Accept equivalent forms.
1 mark
A1 \(v_y^2 = \left(\dfrac{7U}{25}\right)^2 + 784\) oe, where \(v_y\) is the vertical component at \(C\)
g need not be substituted for this mark.
1 mark
M1 Uses the direction of motion at \(C\): \(\tan 60^\circ = \dfrac{v_y}{U\cos\alpha}\), with the unchanged horizontal component \(U\cos\alpha = \dfrac{24U}{25}\), to form an equation in \(U\) only
Independent of the A mark before it. Condone the reciprocal ratio (\(\tan 30^\circ\)) for this mark. M0 if the horizontal component is not resolved, or if the 60° is used as the angle of the line \(AC\) instead of the direction of motion.
1 mark
A1 \(3\left(\dfrac{24U}{25}\right)^2 = \left(\dfrac{7U}{25}\right)^2 + 784\), i.e. \(\dfrac{1679U^2}{625} = 784\) oe
\(U^2 = 291.8\).
1 mark
A1 \(U = 17\) or 17.1
Depends on both previous M marks. awrt 17.1, i.e. in [17.05, 17.15), or 17. Answers from g = 9.8 should be given to 2 or 3 s.f.; a more accurate final answer loses this A mark (over-accuracy is penalised once per question). Use of g = 9.81 is penalised once per question.
1 mark
M1 Finds the time from \(A\) to \(C\) with their \(U\): \(t = \dfrac{U\sin\alpha + v_y}{9.8}\), or solves \(-40 = (U\sin\alpha)t - 4.9t^2\) for the positive root
Their \(U\) and \(v_y\). Accept equivalent forms.
1 mark
A1 \(t = 3.39\) s or 3.4 s
awrt 3.39, i.e. in [3.385, 3.395), or 3.4.
1 mark
M1 Uses horizontal motion: \(BC = (U\cos\alpha) \times \text{their } t\)
The horizontal component must be resolved.
1 mark
A1 \(BC = 56\) m or 55.5 m
awrt 55.5, i.e. in [55.45, 55.55), or 56. Premature rounding (e.g. using \(U = 17.1\), which gives 55.6) loses this mark. Answers from g = 9.8 should be given to 2 or 3 s.f.; a more accurate final answer loses this A mark (over-accuracy is penalised once per question). Use of g = 9.81 is penalised once per question.

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