Projectile motion — horizontal - vertical independence, full derivation: Visual walkthrough

Definition The setup and its assumptions (read first)

Before any symbol appears, fix the stage:

Launch point = origin. We measure everything from where the ball leaves the hand.
$x$ — the horizontal displacement: how far sideways the ball has moved from the launch point, in metres. $x = 0$ at launch.
$y$ — the vertical displacement: how high above (or, if negative, below) the launch point the ball is, in metres. $y = 0$ at launch.
$t$ — the elapsed time since launch, measured in seconds, with $t = 0$ the instant the ball leaves the hand.
Assumption 1 — no air resistance. The only force during flight is gravity.
Assumption 2 — constant $g$ . Gravity's strength does not change over the flight (true for the small heights we deal with).
Assumption 3 — level ground. Launch and landing happen at the same height ( $y = 0$ ). Step 9 shows what to change when this fails (a cliff).

Step 1 — One arrow, two shadows

WHAT we do: We throw a ball from the corner of our page (the "origin", the point where both a horizontal ruler and a vertical ruler read zero). It leaves with a speed $u$ — how fast, in metres per second — pointed at an angle $θ$ measured up from the flat ground.

WHY we do it: A slanted arrow is hard to reason about directly. But a slanted arrow secretly contains two simpler arrows: how much of it points sideways and how much points up. If we can split it, we can handle each piece with plain 1-D motion (see Vectors — Resolving into Components).

PICTURE: Look at the figure. The red arrow is the launch velocity. Drop a straight line from its tip down to the horizontal ruler — the length of that shadow is the horizontal piece. Slide across to the vertical ruler — that shadow is the vertical piece. The arrow, its two shadows, and the corner form a right triangle.

Figure — Projectile motion — horizontal - vertical independence, full derivation

Why cosine for horizontal and sine for vertical? Because on the launch triangle the ground side sits next to (adjacent to) the angle, and the height side sits across from (opposite) it. That is the whole reason — no formula to memorise, just read the triangle.

Step 2 — Two separate stories, one shared clock

WHAT we do: We ask: once the ball is flying, what pushes it? Only gravity, pulling straight down with strength $g$ (about $10 m/s^{2}$ near Earth — see Free Fall and g). There is nothing pushing sideways.

WHY we do it: By Newton's Second Law applied to each direction ( $F_{x} = m a_{x}$ , $F_{y} = m a_{y}$ ), a force that points only downward can only change the up-down motion. The sideways motion has no force on it, so nothing changes it.

PICTURE: The red gravity arrow points purely down. Its shadow on the horizontal ruler has zero length — that is the visual statement "gravity has no sideways part". Two side panels show the two independent stories that share one stopwatch $t$ (the elapsed time since launch).

The only thing linking the two stories is $t$ , the single clock ticking for both.

Step 3 — From acceleration to velocity (one integration)

WHAT we do: Acceleration is the rate at which velocity changes. To go from "how fast velocity changes" back to "the velocity itself" we add up all those tiny changes over the elapsed time $t$ — that adding-up is integration (see Calculus — Integration).

WHY integration and not something else? Acceleration answers "change per second". Velocity is the running total of that change. Undoing a rate to recover a total is exactly what integration does. Nothing else recovers a quantity from its rate.

PICTURE: Two graphs of velocity versus time. The horizontal velocity is a flat red line — it never budges. The vertical velocity is a red line sloping downward — it starts at $u sin θ$ and drops by $g$ each second, crossing zero at the peak, then going negative (falling).

Step 4 — From velocity to position (integrate again)

WHAT we do: Velocity is the rate at which position changes. Add it up over the elapsed time $t$ again and we get where the ball actually is — its horizontal displacement $x$ and vertical displacement $y$ measured from the launch point (both defined in the setup box above).

WHY: Same logic as Step 3, one level up. Position is the running total of velocity, so we integrate a second time. (See 1-D Kinematics — Equations of Motion — these are exactly those equations, one per axis.)

The two integrations, explicitly:

Horizontal velocity $v_{x} = u cos θ$ is constant, so its running total over time $t$ is just constant $\times$ time: $x = (u cos θ) t$ .
Vertical velocity $v_{y} = u sin θ - g t$ has two pieces. Integrating the constant part $u sin θ$ over time gives $(u sin θ) t$ . Integrating the growing part $- g t$ gives $- \frac{1}{2} g t^{2}$ — this is where the $\frac{1}{2}$ comes from: the total of a quantity that grows linearly from $0$ to $g t$ is the average value $\frac{1}{2} g t$ times the time $t$ , i.e. $\frac{1}{2} g t^{2}$ (which is the area of that triangle in the picture).

PICTURE: On the velocity-time graphs, the shaded red area under each line is the distance travelled. The flat line's area is a rectangle → grows steadily. The sloped line's area is a triangle → the classic $- \frac{1}{2} g t^{2}$ falling term.

Everything below is squeezed out of these two lines.

Step 5 — Erase the clock: the parabola appears

WHAT we do: We want the shape of the flight — vertical displacement $y$ as a function of sideways distance $x$ , with time hidden away. So we solve the $x$ -equation for $t$ and substitute it into the $y$ -equation.

WHY: A photograph of the arc has no clock in it. To match that photo we must eliminate $t$ and get a pure $y$ -versus- $x$ relationship.

PICTURE: The red curve is the flight. Notice it has the algebraic form $y = (something) x - (something) x^{2}$ — a term linear in $x$ pulling it up, and a term in $x^{2}$ pulling it down. That is the signature of a downward parabola.

Step 6 — When does it land? Time of flight $T$

WHAT we do: "Landing on level ground" means the height returns to zero, $y = 0$ . We solve $y = 0$ for time.

WHY: Only the vertical story decides when the ball touches down — sideways motion never brings it back to the ground. So we set the $y$ -equation to zero. (This uses Assumption 3 — level ground.)

PICTURE: The parabola crosses the ground line at two red dots: one at the start ( $t = 0$ ) and one at landing ( $t = T$ ). Between them the peak sits exactly halfway.

Step 7 — How high? Maximum height $H$

WHAT we do: At the very top the ball stops rising: its upward speed is momentarily zero. Set $v_{y} = 0$ , find that instant, plug it into $y$ .

WHY: The peak is the one place where $v_{y}$ crosses zero (from up to down). That single condition pinpoints the top.

PICTURE: At the apex the vertical red arrow has shrunk to nothing, while the horizontal red arrow is still full-length — the ball is still cruising sideways even at the very top.

Step 8 — How far? Range $R$

WHAT we do: The ball drifts sideways at constant speed for the whole flight, so range = sideways speed $\times$ total time.

WHY: Horizontal speed never changes (Step 2), so distance is simply speed times time — the easiest multiplication in physics.

PICTURE: A red horizontal bar of length $R$ under the arc. Above it, a small dial shows $R$ swelling to its biggest value exactly at $θ = 4 5^{\circ}$ , then shrinking again.

Step 9 — The edge cases (never skip these)

WHAT we do: We test the derivation at its extreme angles, and we lift Assumption 3 to see what changes.

WHY: A rule you only tested in the "nice" case is not a rule you trust. Check the corners.

PICTURE: Three miniature red flights side by side: fired flat, fired straight up, and the balanced $4 5^{\circ}$ throw.

The one-picture summary

Every step, one diagram: the red arc, its two shadow-motions, the landing dots, the peak, and the range bar — the entire derivation compressed.

Recall Feynman retelling — the whole walkthrough in plain words

Throw a ball. That single slanted push is really two pushes in disguise: a sideways one and an upward one — the two shadows of the arrow. Now let go. Gravity only tugs down, so it never touches the sideways shadow — the ball keeps gliding forward at a perfectly steady pace forever. Meanwhile the upward shadow is being eaten away by gravity: it slows, stops at the top, then falls back faster and faster. Add "steady forward" to "rise-then-fall" and you draw a smooth arch — a parabola. When does it land? When the up-down story returns to the ground — that's the time of flight, and it's set only by the upward shadow. How high? Wherever the upward speed hits zero. How far? Steady sideways speed times the total time — biggest when you split the throw evenly at forty-five degrees. And if you fire it flat off a cliff, it falls in step with a ball you simply drop — clink, together — because their up-down stories are the very same story.

Recall Rebuild it yourself

Which integration gives velocity from acceleration? ::: The first one; velocity is the running total of acceleration. What condition finds the time of flight? ::: Set $y = 0$ (back to launch height). What condition finds the peak? ::: Set $v_{y} = 0$ . Why is the path a parabola? ::: After erasing $t$ , $y$ becomes quadratic in $x$ . Which angle maximises range and why? ::: $4 5^{\circ}$ , because $sin 2 θ = 1$ there.