What are a strategy for projectile problems?

Almost every projectile question is solved by the same plan. First, resolve the launch velocity into a horizontal part ucosθ and a vertical part usinθ. Second, write down the horizontal motion, which is simply constant velocity, so x = ucosθ · t. Third, write down the vertical motion as a suvat problem with acceleration -g.

EnglandMathsSyllabus dot point

How do you analyse the motion of a body launched into the air under gravity?

Projectile motion resolved into horizontal and vertical components, the independence of the two motions, and finding range, maximum height, time of flight and the path.

A focused answer to the Edexcel A-Level Mathematics projectiles content, covering motion resolved into horizontal and vertical components, the independence of the two motions, and finding range, maximum height, time of flight and the equation of the path.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Examples in context
Try this

What this dot point is asking

Edexcel wants you to model projectile motion by resolving the initial velocity into horizontal and vertical components, treat the two directions independently (constant horizontal velocity, vertical acceleration $g$ ), and find the time of flight, range, greatest height and the equation of the trajectory.

The answer

Resolving the velocity

The independent motions

A strategy for projectile problems

Almost every projectile question is solved by the same plan. First, resolve the launch velocity into a horizontal part $u\cos\theta$ and a vertical part $u\sin\theta$ . Second, write down the horizontal motion, which is simply constant velocity, so $x = u\cos\theta \cdot t$ . Third, write down the vertical motion as a suvat problem with acceleration $-g$ . Fourth, use the time $t$ as the bridge between the two: typically you find $t$ from the vertical equation (time to reach the ground or the top), then feed that $t$ into the horizontal equation for the range. Keeping the two columns separate on the page, one headed horizontal and one headed vertical, prevents the most common errors.

Examples in context

A ball is kicked at

25 \text{ m s}^{-1}

40

degrees from ground level. Find the speed and direction of motion after

1.5

s. Take

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

step 1 Components of the initial velocity

$u_x = 25\cos 40 \approx 19.15 \text{ m s}^{-1}$ , $u_y = 25\sin 40 \approx 16.07 \text{ m s}^{-1}$ .

step 2 Velocity components at $t = 1.5$

Horizontal stays constant: $v_x = 19.15 \text{ m s}^{-1}$ . Vertical: $v_y = u_y - gt = 16.07 - 9.8 \times 1.5 = 16.07 - 14.7 = 1.37 \text{ m s}^{-1}$ .

step 3 Combine into speed and direction

Speed $= \sqrt{v_x^2 + v_y^2} = \sqrt{19.15^2 + 1.37^2} \approx \sqrt{366.7 + 1.9} \approx 19.2 \text{ m s}^{-1}$ . The angle above the horizontal is $\arctan\!\left(\dfrac{1.37}{19.15}\right) \approx 4.1$ degrees.

A stone is thrown at

20 \text{ m s}^{-1}

60

degrees from the edge of a cliff and lands on the beach

25

m below. Find the time of flight. Take

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

step 1 Vertical component and equation

$u_y = 20\sin 60 \approx 17.32 \text{ m s}^{-1}$ . Taking up as positive, the landing point is $25$ m below the launch, so $-25 = u_y t - \tfrac{1}{2}gt^2$ .

step 2 Form the quadratic

$-25 = 17.32t - 4.9t^2$ , so $4.9t^2 - 17.32t - 25 = 0$ .

step 3 Solve with the quadratic formula

$t = \dfrac{17.32 + \sqrt{17.32^2 + 4 \times 4.9 \times 25}}{2 \times 4.9} = \dfrac{17.32 + \sqrt{300 + 490}}{9.8} = \dfrac{17.32 + 28.11}{9.8} \approx 4.64$ s, taking the positive root.

Try this

Q1. A particle is projected horizontally at $15$ m per second from a height of $20$ m. Taking $g = 9.8$ , find the time to land. [3 marks]

Cue. $20 = \tfrac{1}{2}(9.8)t^2$ gives $t = \sqrt{\dfrac{40}{9.8}} \approx 2.02$ s.

Q2. For a launch at $25$ m per second at $40$ degrees, find the horizontal component of velocity. [2 marks]

Cue. $25\cos 40 \approx 19.2$ m per second.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20207 marksA particle is projected from a point on horizontal ground with speed

28

m s

^{-1}

at an angle of

30

degrees above the horizontal. Find the time of flight and the horizontal range. Take

g = 9.8

m s

^{-2}

Show worked answer →

Resolve the launch velocity (M1): horizontal $u_x = 28\cos 30 = 28 \times \dfrac{\sqrt{3}}{2} \approx 24.25$ m s $^{-1}$ ; vertical $u_y = 28\sin 30 = 28 \times 0.5 = 14$ m s $^{-1}$ (A1).

For time of flight, the vertical displacement returns to zero: $0 = u_y t - \dfrac{1}{2}gt^2$ (M1), so $t(14 - 4.9t) = 0$ , giving $t = \dfrac{14}{4.9} \approx 2.86$ s (A1).

Horizontal range $R = u_x t = 24.25 \times 2.86 \approx 69.3$ m (M1, A1).

Markers reward resolving, the time-of-flight equation, solving for $t$ , and the range from the constant horizontal velocity. (A1 for accuracy.)

Edexcel 20236 marksA stone is thrown horizontally with speed

12

m s

^{-1}

from the top of a cliff

45

m above the sea. Calculate the time taken to reach the sea and the horizontal distance travelled. Take

g = 9.8

m s

^{-2}

Show worked answer →

The initial vertical velocity is zero, so vertically $45 = \dfrac{1}{2}gt^2$ (M1): $45 = 4.9t^2$ (A1).

Solve: $t^2 = \dfrac{45}{4.9} = 9.184$ , so $t = 3.03$ s (M1, A1).

Horizontally the velocity is constant at $12$ m s $^{-1}$ , so the distance is $x = 12 \times 3.03 \approx 36.4$ m (M1, A1).

Markers reward using zero initial vertical velocity, finding $t$ from the vertical drop, and the horizontal distance.

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Mathematics (9MA0) specification — Pearson Edexcel (2017)