Home › Theme A › A.1 › Concept builder, part 2

A.1 Kinematics · Part 2 of 3 · student investigation · 80 minutes

Falling, flying and drag

Measure g from a strobe photo, solve vertical throws with one sign rule, split projectiles into two motions, and see what air resistance changes.

1 Predict · 6
2 Observe g · 10
3 Throw up · 8
4 From a cliff · 10
5 Sideways · 8
6 At an angle · 14
7 Drag · 10
8 Show it · 8
9 Reflect · 6

Learning goals: by the end you can…

find the acceleration of free fall from position–time data.
solve vertical throws using one sign convention (up positive).
explain why horizontal and vertical motions of a projectile are independent.
solve projectile problems for horizontal and angled launches.
describe the effects of air resistance and explain terminal speed.

How to work in this lesson

How to work. Pairs discuss and agree, but each of you writes. Pairs → individual discuss the method, then write your own answer. Individual no discussion until your teacher says so. Use g = 9.8 m s⁻². Your typed answers are saved on this device only.

Everything you need is on this page: each stage has its own simulation, and some stay locked until you have made your prediction. Extra help after class: the A.1 visual guide. ATL key: Thinking = predict / analyse / explain · Communication = explain / share · Social = collaborate · Self-management = plan / monitor / reflect · Research = observe / use evidence.

1. Start with a prediction

6 minPairs

ATL Thinking: predict and justify. Social: compare with a partner.

Warm-up · 3 minutes · no notes

From the last lesson. Answer from memory, then check.

W1A cart starts from rest with a = 3.0 m s⁻² for 4.0 s. Distance travelled =

W2A cart moving at 6.0 m s⁻¹ speeds up at 2.0 m s⁻² over 16 m. Final velocity =

m s⁻¹

W3An object has v = −4.0 m s⁻¹ and a = −2.0 m s⁻². It is…

slowing down, because a is negative speeding up, because v and a have the same sign at rest

1.1From the same table, ball X rolls off the edge at 3 m s⁻¹ and ball Y is dropped at the same moment. Which lands first? (no air resistance)

A. X, because it is moving faster B. Y, because it goes straight down C. They land at the same time

Commit to an answer. You will test it in stage 5.

Check your answer

C. Both start with zero vertical velocity and both have acceleration g downwards, so they fall the same height in the same time.

1.2A steel ball and a plastic ball of the same size are dropped together (no air resistance). Which lands first? Explain.

Check your answer

They land together. Without air resistance every object has the same acceleration g, whatever its mass. With air, the lighter ball would be slowed more (stage 7).

2. Observe: measure g

10 minPairs

ATL Research: take measurements from data. Thinking: find a pattern.

A strobe photo shows a falling ball every 0.10 s. The distances fallen from the start are in the table. Run the simulation to see the photo being taken.

Simulation · Strobe photo of a falling ball a flash every 0.10 s

time—

distance fallen—

speed now—

t / s	distance fallen / m	average speed in the next 0.10 s / m s⁻¹
0.00	0.000	0.49
0.10	0.049	1.47
0.20	0.196
0.30	0.441
0.40	0.784
0.50	1.225	—

2.1By how much does the average speed increase every 0.10 s? Use it to find the acceleration, in m s⁻².

Back check

2.bCheck 2.1 a second way, with s = ½gt² and the last row of the table (1.225 m at 0.50 s). g = 2s/t² =

m s⁻²

2.2The distance column goes 1 : 4 : 9 : 16 : 25 (× 0.049 m). Which equation explains this pattern?

What a good answer looks like

s = ½gt² (from s = ut + ½at² with u = 0). Distance is proportional to t², so doubling the time gives four times the distance.

3. Explain: one sign rule for a throw

8 minPairs → individual

ATL Thinking: apply a convention consistently. Communication: explain the meaning of a sign.

The ruleTake up as positive. Then a = −9.8 m s⁻² for the whole flight: going up, at the top and coming down. Write every s, u and v with its sign.

A ball is thrown straight up at 14.7 m s⁻¹.

3.0Predict: at the very top of its path, the acceleration of the ball is…

zero, because it stops 9.8 m s⁻² downwards 9.8 m s⁻² upwards

Simulation · Throw it up up is positive

throw speed u 14.7 m s⁻¹time t 0.00 s

height s—

velocity v—

acceleration a—

3.1Time to reach the top =

3.2Maximum height =

3.3Velocity at t = 2.0 s =

m s⁻¹

Back check

3.bCheck 3.2 without v²: use s = ½(u + v)t with u = 14.7, v = 0 at the top and your time from 3.1. s =

3.4What does the minus sign in 3.3 tell you?

What a good answer looks like

The ball is moving downwards at 4.9 m s⁻¹. It passed the top at 1.5 s and has been falling for 0.5 s.

4. Build: a throw from a cliff

10 minIndividual

ATL Thinking: plan a multi-step solution. Self-management: check the sign of every answer.

📘 From the data booklet (A.1)

s = ((u + v)/2)tv = u + ats = ut + ½at²v² = u² + 2as

You do not need to memorize these: they are given in the data booklet in the exam. Your job is to choose the right one. For a throw, use a = −g = −9.8 m s⁻² (up positive).

A stone is thrown straight up at 10 m s⁻¹ from the top of a cliff 25 m above the sea. Origin at the throw, up positive: the sea is at s = −25 m.

4.1Write the equation for the time t to reach the sea, then solve it. t =

Simulation · Check your answer to 4.1: the cliff throw origin at the throw, up positive

time—

position s—

velocity v—

4.2Speed when it hits the sea (use v² = u² + 2as) =

m s⁻¹

Back check

4.bCheck 4.1 and 4.2 together. Put your time from 4.1 into v = u + at (up positive). v at the sea =

m s⁻¹

4.3The quadratic in 4.1 has a negative root too. Why do you ignore it?

What a good answer looks like

A negative time would be before the stone was thrown. The equation does not know that, but the physics does, so only the positive root has meaning.

5. Test your prediction: sideways launch

8 minPairs

ATL Research: use the simulation as evidence. Thinking: split one motion into two.

A ball X rolls off a table 1.25 m high at 3.0 m s⁻¹. At the same moment ball Y is dropped from the same height. Run the simulation to test your stage 1 prediction.

Hinge question

Choose your answer. You will see at once if it is right, and why.

A stone is thrown straight down at 5.0 m s⁻¹ from a bridge. Take up as positive. Which pair of values is correct?

A. u = +5.0 m s⁻¹, a = −9.8 m s⁻² B. u = −5.0 m s⁻¹, a = −9.8 m s⁻² C. u = −5.0 m s⁻¹, a = +9.8 m s⁻² D. u = +5.0 m s⁻¹, a = +9.8 m s⁻²

Simulation · Rolled off and dropped together table height 1.25 m · a photo every 0.10 s

speed of X off the table 3.0 m s⁻¹

time—

height of X—

height of Y—

horizontal distance of X—

5.1Time to reach the floor (vertical motion only) =

5.2Horizontal distance from the table =

5.3Speed on landing (combine vₓ = 3.0 and vᵧ = 9.8 × 0.505) =

m s⁻¹

Back check

5.bCheck 5.3 without finding vᵧ from the time: use vᵧ² = 2gh for the fall of 1.25 m, then v = √(vₓ² + vᵧ²). v =

m s⁻¹

5.4Was your stage 1 prediction right? Explain using the word "independent".

What a good answer looks like

The horizontal and vertical motions are independent: gravity acts only vertically, so the sideways speed does not change how fast the ball falls. Both balls start with zero vertical velocity, so they land together.

6. Apply: launch at an angle

14 minPairs → individual

ATL Thinking: compare two cases and generalise. Communication: explain a pattern.

A ball is launched from level ground at 16 m s⁻¹, 30° above the horizontal.

6.0Predict: at the same speed, which launch angle gives the longest range?

30° 45° 60°

Simulation · Launch at an angle level ground · no air resistance · earlier paths stay as grey traces

launch speed u 16 m s⁻¹angle θ 30°

uₓ—

uᵧ—

max height H—

time of flight T—

range R—

6.1Components: uₓ = m s⁻¹, uᵧ = m s⁻¹

6.2Maximum height =

6.3Time of flight =

6.4Range =

Back check

6.bCheck 6.4 with the range formula R = u² sin 2θ / g. R =

6.5Repeat for 60° (same speed). Maximum height =

Find the mistake

A student wrote:

“uᵧ = 16 cos 30° = 13.9 m s⁻¹, so the time of flight is T = 2uᵧ/g = 2 × 13.9 / 9.8 = 2.83 s.”

What is wrong?

Nothing is wrong. The vertical component uses sin, because the angle is measured from the horizontal: uᵧ = 16 sin 30° = 8.0 m s⁻¹, so T = 1.63 s. The time of flight should be uᵧ/g, not 2uᵧ/g.

6.630° and 60° give the same range. Use the simulation to find the angle with the longest range (try 40°, 45° and 50°), and explain why the pairs match.

What a good answer looks like

R = u² sin 2θ / g. sin 2θ is largest at 2θ = 90°, so 45° gives the longest range. For 30° and 60°, sin 60° = sin 120°, so the ranges are equal: the higher path spends longer in the air but moves more slowly across.

What if? · stretch

6.wAn astronaut repeats the same launch on the Moon (16 m s⁻¹ at 30°, g = 1.6 m s⁻²). Range =

7. What changes with air resistance?

10 minPairs

ATL Thinking: reason qualitatively with forces. Research: test your answers with the simulation.

7.1A skydiver has just jumped. As her speed increases, the drag force…

stays the same increases, so her acceleration decreases increases, so her acceleration increases

7.2At terminal speed her acceleration is…

zero g g upwards

7.3Compared with no air resistance, a ball thrown with air resistance has a range that is…

longer the same shorter

Simulation · Skydiver with air resistance m = 70 kg · drag grows with speed

time—

speed—

drag—

acceleration—

7.4Explain why a parachute gives a lower terminal speed.

The parachute makes the drag ________ at a given speed, so drag equals the weight (in air, buoyancy is negligible) at a ________ speed.

What a good answer looks like

The parachute has a much larger area, so the drag is much larger at a given speed. Drag equals the weight at a much lower speed, so the terminal speed is lower and the landing is safe.

8. Show it

8 minIndividual

ATL Communication + Thinking: choose how to show your understanding.

Success criteria: your answer must…

treat the horizontal and vertical motions separately and link them by time;
use g with the correct sign;
give a numerical answer with a unit, or a clear explanation with a force argument.

Option A · Challenge

A ball is launched horizontally from a table 0.80 m high and must land in a cup 2.0 m away. Find the launch speed.

Option B · Explain

Explain terminal speed to a student in Grade 9, with a labelled sketch of the forces at three moments.

Option C · Sketch

Sketch vₓ–t and vᵧ–t graphs for the 30° launch in stage 6, with values at the start, top and end.

8.1Your answer.

What a good answer looks like

Option A: vertical: 0.80 = ½ × 9.8 × t² → t = 0.404 s. Horizontal: u = 2.0 / 0.404 = 4.95 m s⁻¹.
Option B: at the start only the weight acts, so she speeds up at g; as she speeds up, drag grows and the acceleration falls; when drag = weight (in air, buoyancy is negligible) the resultant force is zero, the acceleration is zero and the speed stays constant: terminal speed.
Option C: vₓ–t is a horizontal line at 13.9 m s⁻¹. vᵧ–t is a straight line from +8.0 m s⁻¹ through 0 at 0.82 s to −8.0 m s⁻¹ at 1.63 s, gradient −9.8 m s⁻².

9. Reflect

6 minIndividual

ATL Self-management: retrieve without notes, then plan your next step.

Exit ticket (no notes)

1A stone is dropped from rest. Distance fallen after 2.0 s, in m =

2A ball is thrown up at 19.6 m s⁻¹. Time to the top, in s =

3Launched horizontally at 5.0 m s⁻¹, in the air for 0.60 s. Horizontal distance in m =

4At the top of a projectile's path, the velocity is…

zero horizontal vertical

5A stone falls through air at terminal speed (buoyancy in air is negligible). The drag on it is…

equal to the weight zero larger than the weight

6Which idea from today is still the least clear to you?

What a good answer looks like

There is no single right answer. A good answer names one idea exactly, for example: "I mix up the signs when the ball lands below the launch point."

7Feedforward: Next time I will ________ before I ________.

What a good answer looks like

A good feedforward names an action: "Next time I will find the time from the vertical motion before I use the horizontal motion."

TOK · 2 minutes

Theory of knowledge: how do we know? There is no single right answer. Write two or three sentences.

TOKYou measured g in class and your value was a little different from 9.81 m s⁻². When a measurement disagrees with an accepted value, how do we decide which one to trust?

Points you could make

Compare the difference with the uncertainty: if the accepted value is inside your error bar, there is no real disagreement.
The accepted value comes from many independent measurements, so one class result does not overturn it; look for systematic errors first.
But accepted values can change when careful, repeated measurements disagree with them.

Back to A.1 Part 1 Part 3 Read the notes