1-3 Projectile Motion at an Angle

Chapter 1 · Mechanics

❓ Lesson question

At the highest point of a ball’s angled projectile motion, is the ball momentarily at rest?

You will learn to

1Explain how a projectile launched at an angle can be resolved into two independent component motions.
2Calculate the time of flight, maximum height and range of a projectile launched at an angle.
3Describe the motion of a projectile at the highest point, including its horizontal and vertical velocity components.

Watch a slow-motion replay of a free kick. The ball leaves the player’s foot, rises, and appears to pause at the highest point of its trajectory before descending toward the goal. Commentators call this “hang time”.

Watch the replay again and focus only on the highest point. Is the ball truly at rest at that instant?

✍️ 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) At the highest point, is the ball’s velocity zero, minimum but non-zero, or unchanged?

(b) The ball takes 1.8 s to reach the highest point — will the return back down to launch height take more, less, or the same time?

Write a line for each before this opens.

🔬 Explore — pause at the highest point

What you’ll do: use a phone's slow-motion video to track a projectile launched at an angle and look for patterns.

⏱️ 12 min🧰 a phone with slow-motion video · a crumpled-paper ball · a doorway with a clear vertical edge · a ruler

1Toss the paper ball at about 45° so it follows a curved path across an open doorway; a partner films it in slow motion against the door edge.
2Step through the video frames near the highest point, one at a time. Against the fixed door edge, watch the two components of motion separately.
3Mark the centre of the ball in each frame. Using the scale, measure the horizontal and vertical distances between consecutive positions.
4Repeat with a steeper launch angle and record what changes.
5Then use the simulation below, which draws the two components as arrows so you can see them without frame-stepping.

🎛️ Simulation — the apex, and what is really happening there

Watch the two arrows. The red horizontal one never changes length for the whole flight. The blue vertical one shrinks to nothing at the top, then grows downward.

Launch speed19.6 m/s

Launch angle30°

Launch height0.0 m

Horizontal component

17.0 m/s

Constant for the entire flight — including at the highest point.

Vertical component now

9.8 m/s

speed right now: 19.6 m/s

Time to top / total

1.00 / 2.00 s

exactly double — symmetric

Max height / range

4.9 / 33.9 m

Both launches land in the same place, from the same speed. The 60° launch is in the air 1.73× longer and climbs far higher — and still gets no further. A steeper kick buys height and hang time, never distance.

Analyse what you collected

(a) For each launch angle, did the horizontal displacement per frame remain approximately constant as the ball passed through the highest point?

(b) How did the vertical motion change as the ball passed through the highest point?

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

Measurement note

Equal frame numbers mean equal time intervals. So comparing the horizontal displacement over the same count of frames is comparing speeds, without ever needing a stopwatch.

In the Explore, the ball continued to move horizontally through the highest point in approximately equal steps. Meanwhile the vertical displacement between successive frames decreased as the ball rose, became approximately zero at the highest point, then reversed direction and increased as the ball fell. You may have predicted the ball would stop completely at the top; the video evidence says otherwise.

Resolving the launch into components

Resolve the launch velocity $v_{0}$ into two components. The horizontal one is $v_{x} = v_{0} cos θ$ , and with air resistance neglected it stays constant for the whole flight. For the vertical component the ball behaves exactly like an upward throw — it starts at $v_{0} sin θ$ and gravity pulls it back:

v_{y} = v_{0} sin θ - g t y = v_{0} sin θ \cdot t - \frac{1}{2} g t^{2}

The two motions are independent — they share only the same clock $t$ .

At the highest point, only the vertical component becomes zero. The horizontal component remains constant, so the ball is never truly at rest.

Symmetry about the highest point

The motion is symmetric about the highest point only for launch and landing at the same height, with no air resistance. When that holds, the time to rise equals the time to fall:

t_{flight} = 2 t_{top}

And since there is no horizontal acceleration, the range is simply the constant horizontal velocity multiplied by the whole flight time:

R = v_{0} cos θ \times t_{flight}

So if a projectile reaches its highest point after 1.2 s, it returns to its initial height after 2.4 s; and if the horizontal displacement at the highest point is 3.0 m, the total range is 6.0 m.

Point

The motion of an object projected at an angle to the horizontal is projectile motion. It follows a parabolic path and is resolved into horizontal and vertical components, treated as two independent linear motions.
Horizontal component: neglecting air resistance the object moves with constant velocity $v_{x} = v_{0} cos θ$ , so $x = v_{0} cos θ \cdot t$ .
Vertical component: the same motion as a vertical upward throw — constant acceleration $- g$ from an initial $v_{0} sin θ$ .
Only for launch and landing at the same height, and with air resistance neglected, is the motion symmetric about the vertical line through the highest point.

📐 Equation box

Horizontal motion

a = 0 ⟶ x = v_{0} cos θ \cdot t

Vertical velocity

v_{y} = v_{0} sin θ - g t

Vertical displacement

y = v_{0} sin θ \cdot t - \frac{1}{2} g t^{2}

Time to the highest point

t_{top} = \frac{v _{0} s i n θ}{g}

Maximum height (level ground)

H = \frac{( v _{0} s i n θ ) ^{2}}{2 g}

Range (level ground)

R = v_{0} cos θ \times t_{flight}

Key terms

Time of flight: — Total launch-to-landing time; on level ground, twice the time to the highest point.
Maximum height: — How high it climbs — set by the vertical component alone.
Range: — Horizontal distance covered in the whole flight.
Complementary angles: — θ and 90° − θ. From the same speed on level ground they give the same range.

🛠️ Worked example

Situation

A small ball is projected from the ground with an initial speed of 19.6 m/s at an angle of 30° above the horizontal, with $g = 9.8 m/s^{2}$ and no air resistance. Determine (a) the time to reach maximum height, and (b) the horizontal range.

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

The same ball is kicked steeper — 9.8 m/s at 60° to the horizontal. Calculate the time to reach maximum height, and the maximum height.

a-d

A small ball is projected from the ground at 19.6 m/s at 60° to the horizontal. Taking

g = 9.8 m/s^{2}

and

3 \approx 1.73

, determine to two significant figures:
(a) the horizontal component of the initial velocity; (b) the time to reach maximum height; (c) the maximum height; (d) the horizontal range.

In a New Context

A firework shell bursts high above the city, ejecting glowing pellets — called stars — in all directions. Each star then moves as an independent projectile, and each experiences the same downward gravitational acceleration $g$ . Neglecting air resistance, every star follows its own parabola relative to the ground.

Because gravity gives every star the same acceleration, it does not change their motion relative to one another. If the stars are ejected at approximately equal speeds in all directions, they form an expanding, approximately spherical pattern. The centre of that pattern follows a projectile path of its own, while the pattern grows but stays round.

➡️ Explain why the expanding pattern of stars remains approximately spherical, even though each individual star follows a parabolic path.

The answer, if you want to check yourself

Gravity subtracts the same amount from every star’s velocity at every instant. A quantity taken equally from everyone changes nobody’s position relative to anyone else. So in a frame falling with the burst centre, the stars simply fly outwards at constant speed in all directions — a growing sphere. Seen from the ground, that whole sphere is itself falling along a parabola. This is lesson 1-1 returning once more: the same motion, two frames, two different-looking descriptions, both correct.

🧪 Nature of Science — the model's author identified its limits

In Two New Sciences (1638) Galileo showed that, in the absence of air resistance, a projectile follows a parabolic path. That is the model used throughout this lesson. In the same work, however, he pointed out that air resistance causes the actual path to deviate from a parabola, especially at high speeds.

Every physical model has a range of validity, and Galileo named the limits of his own. That habit is worth copying: an equation you can state the limits of is an equation you understand. It is also why the simulation on this page hides the complementary-angle trajectory as soon as the launch height leaves the ground — the equal-range rule stops being true there, and drawing it anyway would be a small lie.

Your turn: Suggest one observation from a real football match that the parabolic model — with air resistance neglected — cannot explain.

✅ Lesson question, answered

At the highest point of a ball’s angled projectile motion, is the ball momentarily at rest?

Never truly at rest. The vertical component of the velocity becomes momentarily zero, while the horizontal component remains constant throughout — so at the apex the ball is still travelling at $v_{0} cos θ$ .

The acceleration, meanwhile, is downward for the whole flight and is never zero, not even at the top. Zero velocity in one direction is not zero acceleration.

The ball only appears to pause because its vertical motion slows to zero and reverses. Its horizontal motion never stops for an instant.

A small ball is projected from the ground at 39.2 m/s at 30° to the horizontal.
(i) Determine the vertical component of the initial velocity. (ii) Determine the horizontal component of the velocity after 1.5 s. (iii) Determine the time to reach maximum height. (iv) Determine the horizontal range.

A small ball is projected from the ground at 20 m/s at 60° to the horizontal.
(i) Determine the horizontal component of the initial velocity. (ii) Determine the vertical component of the velocity after 0.50 s.

A small ball is projected from the top of a building 39.2 m above the ground at 19.6 m/s at 30° above the horizontal.
(i) Determine the time to reach maximum height. (ii) Determine the maximum height above the ground. (iii) Determine the time to hit the ground. (iv) Determine the horizontal range.

📝 Exam-style question

A stone is thrown from ground level at 12 m/s at 30° above the horizontal.
(a) State the horizontal component of its velocity at the highest point.
(b) Calculate the time for the stone to return to the ground.

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)

Several of these are the misconceptions this lesson exists to break. Pick carefully.

At the highest point of a projectile launched at an angle, its velocity is:

At the highest point, the acceleration of the projectile is:

A ball is launched at 20 m/s at 30° above the horizontal. The time to reach maximum height is closest to:

From the same launch speed on level ground, which pair of angles gives the same range?

A projectile is launched at 40 m/s at 60° to the horizontal. Its horizontal component is:

Keeping the speed the same, raising the launch angle from 30° to 45° will:

Part 2 — written answer (the other 50%)

One routine works for every question here: resolve into components, use the vertical part for anything involving time or height, and the horizontal part for anything involving distance. Each question has its own notebook page — use it, then compare your working with the model answer.

Level 1 — direct substitution

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

1.1

A ball is launched at 25 m/s at 37° above the horizontal. Determine its horizontal and vertical components.

1.2

A projectile has an initial vertical velocity component of 24.5 m/s. How long does it take to reach its maximum height?

Level 2 — application in a new context

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

2.1

A ball is kicked from level ground at 20 m/s at 30° above the horizontal. Determine (i) the time of flight, (ii) the maximum height, (iii) the range.

2.2

At the top of its flight a projectile is moving at 15 m/s. It was launched at 25 m/s. Determine the launch angle and the maximum height reached.

Level 3 — exam challenge

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

3.1

A ball is launched from level ground at 30 m/s, first at 25° and then at 65°. Show that the two ranges are equal and find them. Which launch spends longer in the air, and by how much?

3.2

A stone is thrown from the top of a 20 m cliff at 15 m/s at 40° above the horizontal, and lands in the sea below. Determine (i) the maximum height above the sea, (ii) the time to hit the water, (iii) the horizontal distance travelled.

Keep reading — this lesson marks itself done automatically.