A Level Math Revision Free diagnostic
Themed maths · 5 November

Bonfire Night maths activities for A Level

A ready-to-teach lesson for the week of 5 November: a 5-minute starter, a 35-minute main activity, an extension and full worked answers. It practises projectiles, kinematics with calculus and series.

Level
A Level Maths (Year 12 and Year 13)
Time
40 minutes, plus a 10-minute extension
Topics
Projectiles (Year 13 mechanics); Constant and variable acceleration; Arithmetic and geometric series
Equipment
The starter is non-calculator. A calculator is needed for the main activity. Take g = 9.8 m s−2.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A14 minThe firework projectile
Main: task B8 minThe rocket motor (Year 13)
Main: task C13 minRemember, remember: series
Extension10 minFast finishers or homework

Starter (5 minutes)

No calculator.

  1. Find the sum of the first 10 terms of 3, 7, 11, …
  2. Find the sum to infinity of 8 + 4 + 2 + …
  3. A spark is thrown straight up at 14 m s−1. How long does it take to reach its highest point? (g = 9.8 m s−2)
  4. Solve 4.9t2 − 19.6t = 0.

Main activity (35 minutes)

Task A: The firework projectile (14 min)

A firework is launched from ground level at 35 m s−1 at 60° above the horizontal. Model it as a particle moving freely under gravity.

  1. Find the time to reach the highest point and the greatest height, each to 3 significant figures.
  2. The firework explodes at its highest point. How far has it travelled horizontally? Give 3 significant figures.
  3. If it did not explode, how far from the launch point would it land? Give 3 significant figures.
  4. For how long is the firework at least 30 m above the ground? Give 3 significant figures.

Task B: The rocket motor (Year 13) (8 min)

A rocket rises vertically from the ground. Its velocity is v = 24t − 3t2 m s−1 for 0 ≤ t ≤ 8.

  1. Find when the acceleration is zero, and the greatest velocity.
  2. Find the height of the rocket when t = 8.

Task C: Remember, remember: series (13 min)

A display fires 5 fireworks in the first minute, 8 in the second, 11 in the third, and so on.

  1. How many fireworks are fired in the 20th minute, and in the first 20 minutes altogether?
  2. The display has 1000 fireworks. After how many whole minutes have they all been fired?
  3. On 5 November one person shares a photo with 5 friends. The next day each of them shares it with 5 new people, and so on. How many people (not counting the first) have received it after 5 rounds?
  4. After how many rounds have more than one million people received it?

Extension (10 minutes)

Projectiles from a height.

  1. Show that the path of the firework in task A is y = √3x − (2/125)x2.
  2. The same firework is launched from the top of a 20 m high hill, at the same speed and angle. How long is it in the air, and how far away horizontally does it land? Give 3 significant figures.

For teachers

Teacher notes and full worked answers
  • Enjoy fireworks at an organised display; this pack is about the maths only.
  • Keep exact values (35 sin 60° = 35√3/2) until the last line: early rounding changes the third significant figure in A (d).
  • Ask the class what the model ignores (air resistance, the rocket’s own thrust) and how each would change the answers.

Starter

  1. 210
    • S10 = 10/2 × (2 × 3 + 9 × 4) = 5 × 42 = 210
  2. 16
    • a = 8, r = ½: 8 ÷ (1 − ½) = 16
  3. 10/7 s ≈ 1.43 s
    • v = u − gt = 0 gives t = 14 ÷ 9.8 = 10/7.
  4. t = 0 or 4
    • 4.9t(t − 4) = 0

Task A: The firework projectile

  1. 3.09 s; 46.9 m
    • Vertical: u = 35 sin 60° = 30.31… m s−1.
    • Time to the top: 30.31 ÷ 9.8 = 3.09 s.
    • Height: (35 sin 60°)2 ÷ (2 × 9.8) = 918.75 ÷ 19.6 = 46.875, so 46.9 m.
  2. 54.1 m
    • Horizontal speed: 35 cos 60° = 17.5 m s−1.
    • 17.5 × 3.0929… = 54.13 m
  3. 108 m
    • Flight time is twice the time to the top: 6.186 s.
    • 17.5 × 6.186 = 108.25…, so 108 m.
  4. 3.71 s
    • 30.311t − 4.9t2 = 30 gives 4.9t2 − 30.311t + 30 = 0.
    • t = 1.237 or 4.949.
    • Time above 30 m = 4.949 − 1.237 = 3.71 s

Task B: The rocket motor (Year 13)

  1. t = 4; 48 m s−1
    • a = dv/dt = 24 − 6t = 0 when t = 4.
    • v(4) = 96 − 48 = 48
  2. 256 m
    • ∫08 (24t − 3t2) dt = [12t2 − t3]08 = 768 − 512 = 256

Task C: Remember, remember: series

  1. 62; 670
    • u20 = 5 + 19 × 3 = 62
    • S20 = 20/2 × (5 + 62) = 670
  2. 25 minutes
    • n/2 × (10 + 3(n − 1)) ≥ 1000 gives 3n2 + 7n − 2000 ≥ 0.
    • n ≥ (−7 + √24 049) ÷ 6 = 24.68…
    • So 25 minutes (S24 = 948 is not enough).
  3. 3905
    • Σr=15 5r = 5(55 − 1) ÷ 4 = 5 × 3124 ÷ 4 = 3905
  4. 9 rounds
    • 5(5n − 1) ÷ 4 > 1 000 000 gives 5n > 800 001.
    • n > log 800 001 ÷ log 5 = 8.45…, so 9 rounds.

Extension

  1. y = √3x − (2/125)x2
    • x = 17.5t, so t = x ÷ 17.5.
    • y = 30.31t − 4.9t2 = (35 sin 60° ÷ 17.5)x − 4.9x2 ÷ 306.25.
    • 35 sin 60° ÷ 17.5 = √3 and 4.9 ÷ 306.25 = 0.016 = 2/125.
  2. 6.79 s; 119 m
    • −20 = 30.311t − 4.9t2 gives 4.9t2 − 30.311t − 20 = 0.
    • t = (30.311 + √(918.75 + 392)) ÷ 9.8 = 6.787…
    • 17.5 × 6.787 = 118.8, so 119 m.

The same answers are in the answers PDF. Every answer was recomputed by computer before publishing.

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