The shape of the lesson
By the end of the lesson, learners can
- Two motions, one flight: Set a launch speed and angle and fire.
- Resolve, then recombine: For a launch speed u at angle θ to the horizontal, split the velocity once at the start, then handle each direction with its own rules.
- Turning a curve into two lines.
- Horizontal: constant velocity, x = (u cosθ) t.
- Vertical: suvat with a = g, using u sinθ.
- Time links the two; find it from the vertical motion.
One picture every idea returns to
Horizontal and vertical motion are independent. Horizontally there is no force (ignoring air resistance), so the velocity is constant: x = (u cosθ) t. Vertically there is gravity, so the motion obeys the suvat equations with a = g: y = (u sinθ) t − ½ g t². Time is the link between the two.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Sideways and down, at once.. A projectile is two simple motions happening together and never interfering: a steady horizontal velocity and a vertical fall under gravity. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Two motions, one flight | Set a launch speed and angle and fire. Project tab 1 of the lab (Intro) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Resolve, then recombine | For a launch speed u at angle θ to the horizontal, split the velocity once at the start, then handle each direction with its own rules. | Pairs |
| 21 to 30 min | Teach and check: Turning a curve into two lines | Turning a curve into two lines. | Frayer Model |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 15 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Turning a curve into two lines, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Frayer Model
Pairs fill a four-box grid for the key term: definition, characteristics, examples and non-examples. The non-examples box is where the misconceptions surface, so give it the most time.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Never mix the components. | Gravity acts only vertically, so it never changes the horizontal velocity; do not apply g to the horizontal motion. For a ball launched horizontally, the time to land depends on the drop height alone, not on how fast it was thrown sideways: a faster throw simply lands further away in the same time. And remember the acceleration at the top of the arc is still g, not zero, even though the vertical velocity there is zero. |
Projectile motion: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Explain, in terms of forces, why the horizontal velocity of a projectile stays constant while its vertical velocity changes, assuming air resistance is negligible.
A stone is thrown horizontally at 8.0 m s⁻¹ from a cliff 20 m high. Taking g = 9.8 m s⁻², find the time to reach the sea and the horizontal distance travelled.
A ball is launched from the ground at 30 m s⁻¹ at 30° above the horizontal. Taking g = 9.8 m s⁻², find its initial horizontal and vertical velocity components and the time of flight.
For the ball in question 3, find the horizontal range on level ground.
A projectile is fired horizontally at 12 m s⁻¹ from a height and, after 1.5 s, has not yet landed. Taking g = 9.8 m s⁻², find the magnitude of its velocity at that instant.
A marksman aims a rifle horizontally and, at the exact moment the bullet leaves the barrel, a target is released from rest from the same height some distance away. Ignoring air resistance, explain why the bullet still hits the falling target.
Answer key and marking notes
- There is no horizontal force, so no horizontal acceleration, and the horizontal velocity is constant. ✓
- Gravity acts vertically downward, giving a constant vertical acceleration g, so the vertical velocity changes throughout. ✓
- Vertical: 20 = ½ × 9.8 × t², so t² = 4.08 and t = 2.0 s. ✓
- Horizontal: x = 8.0 × 2.0 = 16 m. ✓
- Horizontal: 30 cos 30° = 26 m s⁻¹. Vertical: 30 sin 30° = 15 m s⁻¹. ✓
- Time of flight: t = 2u sinθ / g = 2 × 15 / 9.8 = 3.1 s. ✓
- Range = horizontal velocity × time of flight = 26 × 3.1 = 81 m. ✓
- (Equivalently R = u² sin 2θ / g = 30² × sin 60° / 9.8 ≈ 80 m.) ✓
- Horizontal velocity stays at 12 m s⁻¹. ✓
- Vertical velocity = gt = 9.8 × 1.5 = 14.7 m s⁻¹ downward. ✓
- Speed = √(12² + 14.7²) = √(144 + 216) = √360 ≈ 19 m s⁻¹. ✓
- Both the bullet and the target start with zero vertical velocity and fall under the same g, so at any instant they have dropped by the same amount. ✓
- The bullet's horizontal motion carries it to the target's horizontal position in that time, and since both have fallen equally, they meet. ✓
17 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
A coin is flicked horizontally off a bench at the same instant an identical coin is dropped from the bench edge. Ignoring air resistance, they reach the floor:
- Athe flicked coin first
- Bat the same time
- Cthe dropped coin first
- Dat times set by the flick speed
Ignoring air resistance, the horizontal acceleration of a projectile during its flight is:
- Ag downward
- Bg upward
- Cequal to the launch speed
- Dzero
Throughout the flight of a projectile, its vertical acceleration is:
- Ag, directed downward
- Bzero
- Cg, directed upward
- Dequal to its vertical velocity
A projectile is launched at 40 m s⁻¹ at 60° to the horizontal. Its horizontal velocity component is:
- A35 m s⁻¹
- B40 m s⁻¹
- C20 m s⁻¹
- D23 m s⁻¹
For the same 40 m s⁻¹ launch at 60°, the vertical velocity component is about:
- A20 m s⁻¹
- B35 m s⁻¹
- C40 m s⁻¹
- D12 m s⁻¹
A ball is projected horizontally from a height of 5.0 m. Taking g = 10 m s⁻², the time before it lands is:
- A0.5 s
- B2.0 s
- C0.7 s
- D1.0 s
A ball is projected horizontally at 15 m s⁻¹ and is in the air for 1.0 s. Its horizontal range is:
- A15 m
- B7.5 m
- C30 m
- D1.5 m
Ignoring air resistance, the launch angle that gives the greatest range on level ground is:
- A30°
- B60°
- C45°
- D90°
Two projectiles are launched at the same speed from level ground, one at 20° and one at 70°. Their horizontal ranges are:
- Alarger for 20°
- Bequal
- Clarger for 70°
- Dzero for 70°
At the highest point of its arc, the speed of a projectile equals its:
- Avertical velocity component
- Blaunch speed
- Czero
- Dhorizontal velocity component
Ignoring air resistance, the path of a projectile launched at an angle is a:
- Aparabola
- Bcircle
- Cstraight line
- Dhyperbola
For a fixed launch speed, increasing the launch angle above 45° will:
- Aincrease the range
- Bleave the range unchanged
- Creduce the range
- Dreduce the maximum height
A projectile has an initial vertical velocity component of 20 m s⁻¹. Taking g = 10 m s⁻², the time to reach its highest point is:
- A4.0 s
- B2.0 s
- C1.0 s
- D0.5 s
On level ground, the total time of flight of a projectile is:
- Aequal to the time to reach the highest point
- Bhalf the time to reach the highest point
- Cindependent of the launch angle
- Dtwice the time to reach the highest point
Ignoring air resistance, a projectile launched from and landing at the same height has a landing speed that is:
- Aequal to the launch speed
- Bgreater than the launch speed
- Cless than the launch speed
- Dalways zero
A ball is launched at 50 m s⁻¹ at 37° to the horizontal (sin 37° = 0.60). Its initial vertical velocity component is:
- A40 m s⁻¹
- B50 m s⁻¹
- C30 m s⁻¹
- D25 m s⁻¹
For a ball launched horizontally from a fixed height, doubling the launch speed will:
- Adouble both the range and the time of flight
- Bdouble the range but leave the time of flight unchanged
- Chalve the time of flight
- Dleave the range unchanged
Quiz key
Both start with zero vertical velocity and fall under the same g, so they land together.
With no horizontal force there is no horizontal acceleration, so it is zero and the horizontal velocity stays constant.
Gravity gives a constant downward acceleration g for the entire flight.
Horizontal component = 40 cos 60° = 40 × 0.5 = 20 m s⁻¹.
Vertical component = 40 sin 60° ≈ 35 m s⁻¹.
5.0 = ½ × 10 × t², so t² = 1.0 and t = 1.0 s.
Range = 15 × 1.0 = 15 m.
Range is greatest when sin 2θ = 1, that is θ = 45°.
Range depends on sin 2θ, and 20° and 70° are complementary, so sin 40° = sin 140° and the ranges are equal.
At the top the vertical velocity is zero, so the speed equals the (unchanged) horizontal component.
Combining constant horizontal velocity with uniform vertical acceleration gives a parabolic path.
Beyond 45° the range falls (though the maximum height keeps rising).
0 = 20 − 10t, so t = 2.0 s.
By symmetry the time up equals the time down, so the total flight time is twice the time to the top.
By symmetry, the landing speed equals the launch speed when start and finish are at the same height.
Vertical component = 50 sin 37° = 50 × 0.60 = 30 m s⁻¹.
The fall time is set by the height alone, so it is unchanged; the range, being velocity times that time, doubles.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.