Visual walkthrough — Range, max height, time of flight — all derived

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Step 1 — Draw the throw and name what we see

WHAT. A ball leaves the ground moving fast in a slanted direction. Let us give names to the picture so we can talk about it.

• The speed at launch — how fast the ball leaves your hand — we call $u$. It is one number, measured in metres per second.
• The slant — how steeply upward you threw it, measured up from the flat ground — we call $\theta$ (the Greek letter "theta", just a name for the angle).

WHY. Before any maths, we must pin words to marks. $u$ is a length of arrow; $\theta$ is the opening between that arrow and the ground. Everything downstream is built only from these two.

PICTURE. Look at the amber arrow: its length is the speed $u$, and the amber angle between it and the white ground line is $\theta$. Nothing has moved yet — this is the instant of launch.

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Step 2 — Split the one arrow into two honest arrows

WHAT. The slanted arrow is hard to reason about directly. We replace it with two arrows that add up to it: one lying flat (forward) and one standing straight up.

WHY THIS TOOL — why break it up? Because gravity only ever pulls straight down. It has no opinion about forward motion. So if we separate "forward" from "up", gravity touches only the up-part, and the forward-part becomes trivially simple (nothing pushes it). This split is the single trick that makes projectiles solvable. See Vectors — resolving into components.

To get the lengths of the two new arrows we use a right triangle — a triangle with one square corner. Slide the slanted arrow so it becomes the slanted side (hypotenuse) of such a triangle:

• The flat bottom side has length $u\cos \theta$. Here $\cos \theta$ (cosine) means "how much of the slant points forward" — a fraction between $0$ and $1$.
• The vertical side has length $u\sin \theta$. Here $\sin \theta$ (sine) means "how much of the slant points upward".

PICTURE. The amber slant is the hypotenuse; the cyan arrow along the ground is the forward speed $u\cos \theta$; the white vertical arrow is the upward speed $u\sin \theta$. The little square marks the right angle.

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Step 3 — Watch the up-arrow die, then be reborn (vertical motion)

WHAT. Follow only the vertical part now. It starts at $u\sin \theta$ going up, slows, stops at the top, then speeds back down.

WHY. Gravity subtracts a fixed amount of upward speed every second. That fixed amount is $g$ (about $10\text{m/s}$ every second near Earth). See Free Fall & Acceleration due to gravity. Because the change is steady, the vertical velocity is a straight line sloping down through time.

The vertical velocity at any time $t$ (see Equations of Motion (1D kinematics)): $v_y=\underset{\text{starting up-speed}}{\underbrace{u\sin \theta }}-\underset{\text{speed gravity has removed by time }t}{\underbrace{gt}}$

• $u\sin \theta$ — the up-speed we start with.
• $gt$ — how much up-speed gravity has stolen after $t$ seconds.
• The minus sign — gravity removes upward motion.

PICTURE. The graph: a straight amber line of vertical velocity starting positive at $u\sin \theta$, crossing zero (the peak, marked with a cyan dot), then going negative (falling). Time runs along the white axis.

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Step 4 — Find how long the ball floats (time of flight)

WHAT. We want $T$, the total time from launch until the ball is back at ground level.

WHY this method. The vertical position starts at $0$ and must return to $0$. So we ask: for which times is height zero? The height built from the vertical motion is $y=\underset{\text{how far up the throw carries it}}{\underbrace{u\sin \theta t}}-\underset{\text{how far gravity drags it back}}{\underbrace{\frac{1}{2}g{t}^2}}$

Set $y=0$ (on the ground) and factor out the common $t$: $0=t\left(u\sin \theta -\frac{1}{2}gt\right)$

• $t=0$ — the launch instant itself (yes, it is on the ground then too). We throw this root away.
• $u\sin \theta -\frac{1}{2}gt=0$ — solving gives the landing time.

$\boxed{T=\frac{2u\sin \theta }{g}}$

• $u\sin \theta$ — bigger up-throw ⇒ longer flight.
• $g$ in the denominator — stronger gravity ⇒ shorter flight.
• the $2$ — up-trip plus down-trip, exactly the mirror symmetry from Step 3.

PICTURE. The height-versus-time curve: a smooth amber arch starting at $0$, peaking, returning to $0$. The two ground-touch moments $t=0$ and $t=T$ are marked; the cyan bracket labels the full span $T$.

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Step 5 — Let the forward-arrow ride for time $T$ (the range)

WHAT. Now bring back the forward part from Step 2, $u_x=u\cos \theta$. Nothing pushes it sideways, so it moves at this constant speed the entire time the ball is airborne.

WHY. Distance at constant speed is simply $\text{speed}\times \text{time}$. The time available is exactly the float-time $T$ we just found — this $T$ is the only bridge between the vertical world (which set the clock) and the horizontal world (which used the clock).

$R=u_x\cdot T=\underset{\text{steady forward speed}}{\underbrace{u\cos \theta }}\times \underset{\text{time in air, from Step 4}}{\underbrace{\frac{2u\sin \theta }{g}}}=\frac{2u^2\sin \theta \cos \theta }{g}$

• $u\cos \theta$ — forward speed carried the whole flight.
• $\frac{2u\sin \theta }{g}$ — how long that carrying lasts.
• Multiplying gives $u^2$ (speed appears twice) and a $\sin \theta \cos \theta$ pair.

PICTURE. The full parabola drawn over the ground, with a cyan horizontal bar underneath measuring the landing distance $R$. Small clocks/marks show the forward arrow stepping equal distances in equal ticks (constant speed), while the height rises and falls.

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Step 6 — The double-angle tidy-up

WHAT. The clumsy $2\sin \theta \cos \theta$ has a famous shorter name. From Trigonometric identities — double angle: $2\sin \theta \cos \theta =\sin 2\theta$

WHY use it. Two reasons. (1) It is cleaner. (2) It reveals the best angle instantly: a single sine is easiest to maximise. Substituting:

$\boxed{R=\frac{u^2\sin 2\theta }{g}}$

• $u^2$ — throw twice as fast, land four times as far.
• $\sin 2\theta$ — the whole angle-dependence lives here, and $\sin$ never exceeds $1$.
• $g$ — heavier gravity pulls the landing point in.

PICTURE. A plot of $R$ against launch angle $\theta$ from $0^{\circ }$ to $90^{\circ }$: an amber hump peaking dead-centre at $45^{\circ }$. Cyan guide lines drop from the peak to the axis; the mirror pair (e.g. $30^{\circ }$ and $60^{\circ }$) is marked at equal heights.

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Step 7 — The edge cases (never let the reader fall off the map)

WHAT & WHY. A formula you trust must survive its extremes. We test the corners.

Case $\theta =0^{\circ }$ (thrown flat along the ground). Then $\sin 0^{\circ }=0$, so $R=0$. Correct: with no upward part, the vertical motion gives $T=\frac{2u\sin 0}{g}=0$ — the ball is already on the ground and never leaves. Zero time in air ⇒ zero distance.

Case $\theta =90^{\circ }$ (thrown straight up). Then $\sin 2\theta =\sin 180^{\circ }=0$, so $R=0$. Correct: it goes up and comes straight back down onto your head. Longest flight time, but zero forward travel because $u\cos 90^{\circ }=0$.

Between the extremes. $R$ rises from $0$, peaks at $45^{\circ }$, and falls back to $0$ — the full hump of Step 6. No angle is left unexplained.

Warning — same-height only. Every box here assumed launch and landing at the same level. Throw off a cliff and $y$ does not return to $0$; you must re-solve Step 4's quadratic with the drop included.

PICTURE. Three tiny trajectories side by side: the flat $\theta =0$ streak that never rises, the vertical $\theta =90^{\circ }$ spike that returns to the launch point, and the graceful $45^{\circ }$ arch reaching farthest — all sharing one baseline.

Recall Quick self-test

Why is range zero at both $0^{\circ }$ and $90^{\circ }$? ::: At $0^{\circ }$ there's no air time ($T=0$); at $90^{\circ }$ there's no forward speed ($u\cos 90^{\circ }=0$). Both drivers of range hit zero at the ends. Which two things multiply to give range? ::: Constant forward speed $u\cos \theta$ times air time $T=\frac{2u\sin \theta }{g}$.

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The one-picture summary

PICTURE. Everything on one blueprint: the launch arrow split into $u\cos \theta$ (cyan, flat) and $u\sin \theta$ (white, up); the vertical part driving the clock $T=\frac{2u\sin \theta }{g}$; the forward part riding that clock to trace the amber parabola; the landing distance $R=\frac{u^2\sin 2\theta }{g}$ bracketed below; the peak height $H$ labelled at the top. One glance, one derivation.

Recall Feynman retelling — the whole walk in plain words (click)

You throw a ball at slant. First trick: cut the throw into two — one part running forward, one part shooting straight up. Gravity is lazy: it only fights the up-part. So the up-part is a little story of its own — it climbs, slows, stops, and falls, taking a certain amount of time $T$. Because the slow-down is perfectly steady, the time going up equals the time coming down, which is where the "times two" comes from: $T=\frac{2u\sin \theta }{g}$.

Meanwhile the forward part just cruises — nothing slows it — at a fixed speed $u\cos \theta$. So the distance it lands away is simply cruise speed × float time. Multiply them and you get $\frac{2u^2\sin \theta \cos \theta }{g}$, which trig lets us rename $\frac{u^2\sin 2\theta }{g}$.

Since $\sin$ is biggest at $90^{\circ }$ inside, and that $90^{\circ }$ is twice the launch angle, the best throw is $45^{\circ }$ — a fair balance between staying up long enough and driving forward fast. Throw flatter and you're fast but land too soon; throw steeper and you float long but barely move forward. And at the two extremes — dead flat or straight up — you go nowhere at all.

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Connections

• Range, max height, time of flight — all derived — the parent, where the boxed results live.
• Vectors — resolving into components — the arrow-splitting of Step 2.
• Projectile Motion — components & independence — why forward and up motions don't talk.
• Equations of Motion (1D kinematics) — the $v_y$ and $y$ formulas used in Steps 3–4.
• Free Fall & Acceleration due to gravity — where $g$ comes from.
• Trigonometric identities — double angle — the Step 6 identity.
• Energy method — max height via conservation — an alternate route to $H$.