What is M7.5: Projectiles?

In the standard projectile model, the only acceleration is gravity vertically downwards, so horizontal velocity is constant while vertical velocity changes uniformly.

How many marks are M7.5 questions worth?

M7.5 practice drills on this page are worth 2–8 marks each; marks in a real exam vary by question.

Which past exam papers has M7.5 appeared in?

We have not yet indexed a verified past-paper appearance for M7.5 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

M7.5

Model motion under gravity in a vertical plane using vectors; projectiles.

Draft — not yet indexed

Projectiles

Worked answers and methods for M7.5 on Edexcel A-level Maths 9MA0.

Explanation

In the standard projectile model, the only acceleration is gravity vertically downwards, so horizontal velocity is constant while vertical velocity changes uniformly.
Resolve the initial velocity into horizontal and vertical components, then apply constant-acceleration equations separately with a clearly stated positive direction.
With launch speed $u$ at angle $\theta$ from level ground, eliminating $t$ from $x=u\cos\theta\,t$ and $y=u\sin\theta\,t-\dfrac12gt^2$ gives the equation of the path.
A common error is to use $v=0$ for the whole velocity at greatest height; only the vertical component is zero there, unless the projectile was launched vertically.

A projectile is resolved into horizontal

u\cos\theta

and vertical

u\sin\theta

components.

Worked example

A projectile is launched from level ground at $20\,\text{m s}^{-1}$ at $30^\circ$ above the horizontal. It lands at the same level. Using $g=9.8\,\text{m s}^{-2}$ , find its time of flight, horizontal range and greatest height.

1.The components are $u_x=20\cos30^\circ=10\sqrt3$ and $u_y=20\sin30^\circ=10$ .
2.From vertical displacement $0=10t-4.9t^2$ , the non-zero time is $t=10/4.9=2.0408\,\text{s}$ .
3.The range is $10\sqrt3(2.0408)=35.3\,\text{m}$ .
4.At the top, $0=10^2-2(9.8)h$ , so $h=100/19.6=5.10\,\text{m}$ .

Answer: Time of flight $=2.04\,\text{s}$ ; Range $=35.3\,\text{m}$ ; Greatest height $=5.10\,\text{m}$

Common mistakes

Don't use a same-level horizontal-range formula for a projectile that lands at a different vertical level.
Don't use the full launch speed in both directions instead of resolving it into horizontal and vertical components.

Exam tip

Resolve the initial velocity first, use zero vertical displacement for same-level flight, and keep horizontal acceleration zero.

Worked practice

Tier 1 · Easy

A particle is projected vertically upwards at

14.7\,\text{m s}^{-1}

. Using

g=9.8\,\text{m s}^{-2}

, find the time taken to reach its greatest height and the height gained.

(4)

(Total for Question 1 is 4 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	Time $=1.5\,\text{s}$ Height gained $=11.025\,\text{m}$	4
Notes
Taking upwards as positive, $v=u-gt$ . At greatest height $v=0$ , so $0=14.7-9.8t$ and $t=1.5\,\text{s}$ . Then $v^2=u^2+2as$ gives $0=14.7^2-2(9.8)s$ , so $s=11.025\,\text{m}$ .

(4 marks)

Tier 2 · Standard

A ball is projected horizontally at

12\,\text{m s}^{-1}

from a point

25\,\text{m}

above level ground. Using

g=9.8\,\text{m s}^{-2}

and ignoring air resistance, find the time taken to reach the ground and the horizontal distance travelled, each to 3 significant figures.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Time $=2.26\,\text{s}$ Horizontal distance $=27.1\,\text{m}$	4
Notes
Vertically the initial velocity is zero, so $25=\tfrac12(9.8)t^2$ . Hence $t=\sqrt{50/9.8}=2.2587\ldots\,\text{s}$ , which is $2.26\,\text{s}$ to 3 significant figures. Horizontal velocity remains $12\,\text{m s}^{-1}$ , so the horizontal distance is $12t=27.105\ldots\,\text{m}$ , or $27.1\,\text{m}$ to 3 significant figures. Keep the unrounded $t$ on your calculator for that multiplication rather than retyping a rounded value.

(4 marks)

Tier 3 · Hard

From a point

5\,\text{m}

above horizontal ground, a projectile is launched with velocity

(12\mathbf{i}+5\mathbf{j})\,\text{m s}^{-1}

. Take

\mathbf{j}

vertically upwards and use

g=10\,\text{m s}^{-2}

. Find the Cartesian equation of its path, the horizontal distance to its first impact with the ground, and its speed on impact.

(8)

(Total for Question 3 is 8 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$y=5+\dfrac{5}{12}x-\dfrac{5}{144}x^2$ Horizontal distance $=6(1+\sqrt5)\,\text{m}$ Impact speed $=\sqrt{269}\,\text{m s}^{-1}$	8
Notes
$x=12t$ and $y=5+5t-5t^2$ . Substituting $t=x/12$ gives $y=5+5x/12-5x^2/144$ . At impact, $5+5t-5t^2=0$ , so $t=(1+\sqrt5)/2$ and $x=12t=6(1+\sqrt5)\,\text{m}$ . The impact velocity is $12\mathbf{i}+(5-10t)\mathbf{j}=12\mathbf{i}-5\sqrt5\mathbf{j}$ , so its speed is $\sqrt{12^2+(5\sqrt5)^2}=\sqrt{269}\,\text{m s}^{-1}$ .

(8 marks)

Tier 1 · Easy

A particle is projected at

17\,\text{m s}^{-1}

at an angle

\theta

above the horizontal, where

\sin\theta=\dfrac{8}{17}

. Find the horizontal and vertical components of its initial velocity.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	Horizontal component $=15\,\text{m s}^{-1}$ Vertical component $=8\,\text{m s}^{-1}$ upwards	2
Notes
Since $\theta$ is acute, $\cos\theta=15/17$ . Therefore the components are $17\cos\theta=15\,\text{m s}^{-1}$ horizontally and $17\sin\theta=8\,\text{m s}^{-1}$ vertically upwards.

(2 marks)

Tier 2 · Standard

A particle is projected from a point in a vertical plane. After

1.5\,\text{s}

its velocity is

(14\mathbf{i}-2.7\mathbf{j})\,\text{m s}^{-1}

, where

\mathbf{i}

and

\mathbf{j}

are horizontal and vertically upward unit vectors respectively. Assume that air resistance is negligible and take the downward acceleration as

9.8\,\text{m s}^{-2}

. Find its initial speed and its angle of projection above the horizontal. Give both answers to 3 significant figures.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	Initial speed $=18.4\,\text{m s}^{-1}$ Angle of projection $=40.6^\circ$	4
Notes
Horizontal velocity is constant, so the initial horizontal component is $14$ . Vertically, $-2.7=u_y-9.8(1.5)$ , giving $u_y=12$ . Hence the initial speed is $\sqrt{14^2+12^2}=2\sqrt{85}=18.4\,\text{m s}^{-1}$ to 3 significant figures. The angle is $\tan^{-1}(12/14)=40.6^\circ$ to 3 significant figures.

(4 marks)

Tier 3 · Hard

A particle is projected from horizontal ground with velocity

(18\mathbf{i}+24\mathbf{j})\,\text{m s}^{-1}

, where

\mathbf{i}

and

\mathbf{j}

are horizontal and vertically upward unit vectors respectively. Assume that air resistance is negligible and use a constant downward acceleration of

9.8\,\text{m s}^{-2}

. The particle passes twice through the horizontal level

20\,\text{m}

above the point of projection. Find the exact horizontal distance between these two positions and the exact speed at the later position.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Horizontal distance $=\dfrac{360\sqrt{46}}{49}\,\text{m}$ Later speed $=2\sqrt{127}\,\text{m s}^{-1}$	6
Notes
The vertical displacement is $y=24t-4.9t^2$ . Setting $y=20$ gives $9.8t^2-48t+40=0$ , whose roots differ by $\sqrt{48^2-4(9.8)(40)}/9.8=2\sqrt{184}/9.8=20\sqrt{46}/49$ . Horizontal speed is constant at $18\,\text{m s}^{-1}$ , so the horizontal separation is $18(20\sqrt{46}/49)=360\sqrt{46}/49\,\text{m}$ . At the later crossing the vertical velocity is $-\sqrt{184}=-2\sqrt{46}\,\text{m s}^{-1}$ , so the speed is $\sqrt{18^2+(2\sqrt{46})^2}=\sqrt{508}=2\sqrt{127}\,\text{m s}^{-1}$ .

(6 marks)

Tier 2 · Standard

A particle is projected with velocity

(15\mathbf{i}+12\mathbf{j})\,\text{m s}^{-1}

, where

\mathbf{i}

is horizontal and

\mathbf{j}

is vertically upwards. A vertical wall is

30\,\text{m}

horizontally from the point of projection. Model the particle as moving freely under gravity and ignore air resistance. Take

g=9.8\,\text{m s}^{-2}

. Find the height of the particle above the point of projection when it reaches the wall and find its velocity at that instant.

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	Height $=4.4\,\text{m}$ Velocity $=(15\mathbf{i}-7.6\mathbf{j})\,\text{m s}^{-1}$	5
Notes
Horizontal motion gives $30=15t$ , so the particle reaches the wall when $t=2\,\text{s}$ . Its vertical displacement is $12(2)-\tfrac12(9.8)(2^2)=4.4\,\text{m}$ . The horizontal velocity remains $15\,\text{m s}^{-1}$ and the vertical velocity is $12-9.8(2)=-7.6\,\text{m s}^{-1}$ , giving the stated velocity vector.

(5 marks)

Tier 3 · Hard

A particle is projected from horizontal ground and passes through the point

13\,\text{m}

horizontally and

16.1\,\text{m}

vertically above the point of projection after

1\,\text{s}

. Model the particle as moving freely under gravity and ignore air resistance. Take

g=9.8\,\text{m s}^{-2}

. Find its initial velocity. Hence find its greatest height and its horizontal distance from the point of projection when it returns to the ground, giving the distance to 3 significant figures. Use unrounded values in your working.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	Initial velocity $=(13\mathbf{i}+21\mathbf{j})\,\text{m s}^{-1}$ Greatest height $=22.5\,\text{m}$ Horizontal distance $=55.7\,\text{m}$	6
Notes
Horizontal motion gives $13=u_x(1)$ , so $u_x=13\,\text{m s}^{-1}$ . Vertically, $16.1=u_y(1)-\tfrac12(9.8)(1^2)$ , so $u_y=21\,\text{m s}^{-1}$ . At greatest height, $0=21^2-2(9.8)H$ , giving $H=22.5\,\text{m}$ . The non-zero solution of $0=21t-4.9t^2$ is $t=30/7\,\text{s}$ , so the horizontal distance is $13(30/7)=390/7=55.714\ldots\,\text{m}$ , which is $55.7\,\text{m}$ to 3 significant figures.

(6 marks)

Tier 3 · Hard

A particle is projected from a point with speed

14\,\text{m s}^{-1}

at an angle

\theta

above the horizontal, where

0<\theta<90^\circ

. It passes through a point

20\,\text{m}

horizontally from and

5\,\text{m}

below the point of projection. Model the particle as moving freely under gravity, ignore air resistance and take

g=9.8\,\text{m s}^{-2}

. Show that

2\tan^2\theta-4\tan\theta+1=0

. Hence find all possible exact values of

\theta

in the stated interval. Give all numerical answers exactly.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$2\tan^2\theta-4\tan\theta+1=0$ $\theta=\tan^{-1}(1-\dfrac{\sqrt2}{2})$ or $\theta=\tan^{-1}(1+\dfrac{\sqrt2}{2})$	6
Notes
Write $x=14\cos\theta\,t$ and $y=14\sin\theta\,t-4.9t^2$ . Eliminating $t$ gives $y=x\tan\theta-[9.8x^2/(2(14)^2)]\sec^2\theta$ . At $x=20$ the coefficient is $10$ , so $-5=20T-10(1+T^2)$ where $T=\tan\theta$ . This simplifies to $2T^2-4T+1=0$ , giving $T=1\pm\sqrt2/2$ . Both values are positive, so both inverse-tangent values lie in the stated interval.

(6 marks)

Q10

Tier 3 · Hard

10.

The vectors

\mathbf{i}

and

\mathbf{j}

are horizontal and vertically upward unit vectors respectively. A particle is projected from the origin with velocity

(8\mathbf{i}+15\mathbf{j})\,\text{m s}^{-1}

. It first strikes a plane whose cross-section has equation

y=x/2

. Model the particle as moving freely under gravity, ignore air resistance and take

g=9.8\,\text{m s}^{-2}

. Find the exact time of impact, the distance measured along the plane from the origin to the impact point, and the exact speed of impact. Give all numerical answers exactly.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Impact time $=\dfrac{110}{49}\,\text{s}$ Distance along the plane $=\dfrac{440\sqrt5}{49}\,\text{m}$ Impact speed $=\sqrt{113}\,\text{m s}^{-1}$	6
Notes
$x=8t$ and $y=15t-4.9t^2$ . At the plane, $15t-4.9t^2=(8t)/2=4t$ . Besides $t=0$ , this gives $t=11/4.9=110/49\,\text{s}$ . The impact coordinates are $(880/49,440/49)$ , so the distance from the origin along the plane is $\sqrt{(880/49)^2+(440/49)^2}=440\sqrt5/49\,\text{m}$ . The impact velocity is $8\mathbf{i}+[15-9.8(110/49)]\mathbf{j}=8\mathbf{i}-7\mathbf{j}$ , with speed $\sqrt{113}\,\text{m s}^{-1}$ .

(6 marks)

Verified exam appearances

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Other points in M7 Kinematics

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