1-2 Horizontal Projectile Motion

Chapter 1 · Mechanics

❓ Lesson question

When a coin is projected horizontally off a table while another drops beside it, which hits the floor first — and what determines the result?

You will learn to

1Explain why the horizontal and vertical motions of a projectile occur independently.
2Calculate the flight time and the range of a horizontally launched projectile.
3Determine the speed of a projectile at any instant by recombining its components.

Place two coins at the very edge of a table. Flick one horizontally off the table and, at the same instant, nudge the other just over the edge so that it drops straight down with no horizontal push.

The flicked coin lands some distance away; the other lands almost directly below the edge. Two very different paths — one long and curved, one short and straight.

✍️ Predict before you continue

Write your answer before reading on. A prediction you got wrong and then examined teaches you more than an answer you were handed.

(a) Which coin hits the floor first — the flicked one, the nudged one, or neither?

(b) If you flick the first coin twice as hard, does it stay in the air for a longer time, a shorter time, or the same time?

Write a line for each before this opens.

🔬 Explore — compare the motion of two coins

What you’ll do: launch everyday objects, time their motion, and identify a pattern.

⏱️ 10 min🧰 two identical coins · a flat ruler · a pencil · a quiet room

1Place a flat ruler on a table with one end projecting beyond the edge, pivoted on a pencil. Put one coin on the table beside the ruler and a second coin on the projecting end.
2Flick the projecting end sharply. The ruler launches the first coin horizontally while sliding out from beneath the second, letting it fall vertically. Both begin moving at about the same instant.
3Listen: one strike on the floor, or two? Record what you hear.
4Repeat five times, from a gentle flick to the hardest you can — and record each trial.
5Then use the simulation below, which shows what your ears cannot: whether the two coins are level with each other on the way down, not merely at the end.

🎛️ Simulation — flicked coin vs dropped coin

The dashed grey line joins the two coins. Watch it stay horizontal the whole way down — they are level at every instant, not just at the finish.

Launch speed (the flick)14.7 m/s

Height of the table19.6 m

Fall time

2.00 s

Drag the flick as hard as you like — this does not move. It is √(2h/g); the launch speed is not in it.

Horizontal range

29.4 m

this is what the flick changes

Vertical speed now

0.0 m/s

same for both coins, always

Impact speed

24.5 m/s

√(v₀² + (gt)²)

Look at the red dots, dropped at equal time intervals. Sideways the gaps are equal — nothing pushes or slows the coin horizontally. Downwards the gaps grow — gravity is still pulling. Two motions that ignore each other, in one picture. That is why the path is a parabola, and why the blue dots line up level with the red ones every time.

Analyse what you collected

(a) Did the landing ever split into two clicks as the flick became harder? What pattern does the column show?

(b) The flicked coin travels a much longer, curved path, but the strikes stay together. Why does the longer path not increase the time taken to reach the floor?

✍️ Write your explanation in two lines before you read on.

Measurement note

Two clicks closer together than about 20 ms are heard as a single sound. So a “single click” is evidence that the times agree to within 20 ms — good evidence, but not infinite precision. Repeat in a quiet room, and trust five trials over one.

In the Explore, the two coins struck the floor at nearly the same time even when the launched coin was given very different initial speeds. Increasing its horizontal speed increased the horizontal distance travelled but did not change its time of fall. You may have predicted that the faster coin would land later; the observations did not support that.

Two motions combined in one body

An object projected horizontally follows a parabolic path, and it has two independent motions:

Horizontally, no force acts on it (air resistance ignored), so it keeps a constant velocity: $x = v_{0} t$ .
Vertically, gravity pulls exactly as it pulls any dropped object, so the object is in free fall: $v = g t$ and $y = \frac{1}{2} g t^{2}$ .

The two component motions are independent. They share only one thing — the same clock, the same $t$ .

The parabola, and what sets the time

Combining the components produces the parabolic path, but each part can still be analysed separately. Rearranging the vertical equation gives the flight time:

y = \frac{1}{2} g t^{2} ⟹ t = \frac{2 y}{g}

Read that equation carefully. It contains the height and the strength of gravity — and nothing else. The initial speed $v_{0}$ is absent, which is precisely why flicking harder cannot keep the coin up longer. Once you have $t$ , the range follows from the horizontal motion, $x = v_{0} t$ .

Point

The motion of an object thrown horizontally is called horizontal projectile motion. It is resolved into horizontal and vertical components and treated as two independent linear motions, which share only the same time $t$ .
Horizontal component: the object keeps the constant velocity $v_{0}$ , so the horizontal displacement is $x = v_{0} t$ .
Vertical component: the object is in free fall, so $v = g t$ and $y = \frac{1}{2} g t^{2}$ .
The speed at any instant recombines the two: $V = v_{x}^{2} + v_{y}^{2}$ , with $v_{x} = v_{0}$ always.

📐 Equation box

Horizontal motion (uniform)

x = v_{0} t (a = 0)

Vertical motion (free fall from rest)

v = g t y = \frac{1}{2} g t^{2}

Flight time from a height y

t = \frac{2 y}{g}

Speed at any instant

V = v_{x}^{2} + v_{y}^{2}

Key terms

Projectile: — A body in flight governed by gravity alone, once air resistance is neglected.
Range: — The horizontal distance covered during the flight — set by the launch speed.
Flight time: — How long the fall lasts — set by the height alone, not by the launch speed.
Free fall: — Motion under gravity with no initial vertical velocity and no other force.

🛠️ Worked example

Situation

An object is thrown horizontally at 14.7 m/s from a height of 19.6 m above the ground. Taking $g = 9.8 m/s^{2}$ , determine (a) the time taken to reach the ground, (b) the horizontal range, and (c) the speed just before impact.

Step 1 of 6 — try it yourself before you open the next one.

An object projected horizontally at a speed of 4.0 m/s from a certain height reaches the ground in 2.0 s. Calculate the initial height of the object and its horizontal range.

An object is projected horizontally at a speed of 19.6 m/s from a height of 58.8 m above the ground.
(i) Determine the time taken to reach the ground. (ii) Determine the horizontal range. (iii) Determine the speed just before it hits the ground.

An object projected horizontally at a speed of 3.0 m/s from a certain height reaches the ground in 3.0 s. (i) Determine the initial height. (ii) Determine the horizontal range.

In a New Context

A relief aircraft flies level at 60 m/s, 80 m above a flood-cut village, free-dropping double-bagged grain sacks onto a marked field. The pilot cannot release over the mark.

Each sack leaves the aircraft with the aircraft’s forward velocity, keeps that velocity horizontally (air resistance aside), and falls vertically under gravity meanwhile. The fall takes:

t = \frac{2 \times 80}{9.8} \approx 4.0 s

in which the sack moves forward $60 \times 4.0 = 240 m$ . So the release must happen about 240 m before the aircraft reaches the target.

➡️ Your turn: If the aircraft keeps its constant horizontal velocity, the sack lands almost directly below the aircraft — yet far ahead of where it was released. Explain how both statements are true at once.

The answer, if you want to check yourself

Both are true because the sack never loses its forward speed. It travels forward at 60 m/s for the whole 4.0 s, and so does the aircraft — so they stay level with each other the entire time, exactly like the two coins staying level on the way down. “Far ahead of the release point” and “directly below the aircraft” describe the same 240 m from two different starting places.

🧪 Nature of Science — slowing gravity down so the motion could be timed

Free fall was far too fast for Galileo to time with the clocks he had. So he slowed it down: he rolled balls along smooth, gently inclined grooves and timed them with a water clock. In Two New Sciences (1638) he showed that, ignoring air resistance, a projectile combines uniform horizontal motion with vertical motion under gravity — and that the path is a parabola.

Your two-coin experiment is the same idea reached from the other direction: rather than slowing the motion down, you removed the need to measure time at all by comparing two falls against each other. A single click is a measurement.

Your turn: Explain why studying motion down a gentle, frictionless incline is a fair way of studying vertical free fall.

✅ Lesson question, answered

When a coin is projected horizontally off a table while another drops beside it, which hits the floor first — and what determines the result?

Neither arrives first: they strike together. When air resistance is neglected, gravity governs the vertical motion entirely independently of the horizontal motion.

Increasing the horizontal speed increases the horizontal distance travelled, but it does not change the time of flight from the same height, because that time is $t = 2 y / g$ — a formula in which the launch speed simply does not appear.

Height sets the time. Horizontal speed sets the range. They do not trade with each other — which is the same sentence you met in 1-1 about the ferry and the current.

An object is projected horizontally at a speed of 9.8 m/s from a height of 14.7 m above the ground.
(i) Determine the time taken to reach the ground. (ii) Determine the horizontal range. (iii) Determine the speed just before it hits the ground.

An object projected horizontally at a speed of 6.0 m/s from a certain height reaches the ground in 2.0 s. (i) Determine the initial height. (ii) Determine the horizontal range.

An object is projected horizontally from a height of 0.40 m and lands at point

A

, a horizontal distance of 0.40 m away.
(i) Determine the time from launch until it reaches the ground. (ii) Determine the initial speed. (iii) Determine the height required to make it land at point

B

, 0.80 m away, at the same initial speed.

📝 Exam-style question

A ball rolls off a horizontal bench of height 1.25 m with a speed of 2.0 m/s.
(a) Calculate the time the ball takes to reach the floor.
(b) Determine the horizontal distance from the bench at which the ball lands.

How the real paper is built

The ministry sets 50% multiple choice and 50% written answer. They test different things: the MCQ half rewards eliminating fast and spotting a distractor built from a plausible wrong method; the written half asks you to show that you understand why. Practise both halves — being fluent at one and weak at the other costs you exactly half the paper.

Part 1 — multiple choice (50% of the paper)

Every wrong option is a real mistake someone makes. If you pick one, the explanation names it.

A ball rolls off a table 20 m high at 5.0 m/s. The time it takes to reach the floor is closest to:

Two balls leave a table at the same instant: one is dropped, the other thrown horizontally at 10 m/s. Which lands first?

A projectile is launched horizontally at 12 m/s from a height of 45 m. Its horizontal velocity just before landing is:

The horizontal range of a horizontally launched projectile is exactly doubled by:

An object projected horizontally at 20 m/s falls for 3.0 s. Its horizontal displacement is:

A stone thrown horizontally from a cliff lands 40 m from the base after 4.0 s. Its launch speed was:

Part 2 — written answer (the other 50%)

Every one of these is solved the same way: get the time from the vertical motion first, then use it everywhere else. If you can say that sentence without looking, you have the lesson. Use the notebook button on each question — your working is saved on its own page, so you can set it beside the model answer afterwards.

Level 1 — direct substitution

One step. You are given the formula; put the numbers in.

1.1

A ball rolls off a shelf 1.8 m above the floor. How long before it lands?

1.2

A stone is thrown horizontally at 12 m/s and stays in the air for 1.5 s. How far horizontally does it travel?

Level 2 — application in a new context

Two steps, or you must choose the right component before you can start.

2.1

A marble leaves a bench 0.80 m high at 2.5 m/s. Determine (i) the fall time, (ii) the range, (iii) the speed at which it lands.

2.2

A ball thrown horizontally from a cliff 45 m high lands 60 m from the base. Determine (i) the fall time and (ii) the launch speed.

Level 3 — exam challenge

Multi-step, or a case the textbook does not set and examiners do.

3.1

Two balls leave the same 5.0 m ledge horizontally, one at 4.0 m/s and one at 8.0 m/s.
(i) Which lands first? (ii) How much further out does the faster one land? (iii) Find each ball’s landing speed.

3.2

A helicopter flies level at 40 m/s, 125 m above the ground, and must drop a package onto a marker.
(i) How far before the marker must it release? (ii) Where is the helicopter at the moment the package lands?

Keep reading — this lesson marks itself done automatically.