Range, max height, time of flight — all derived: Worked examples

The scenario matrix

Every projectile question is really one of the cells below. The right column names the example that clears it.

#	Case class	What makes it tricky	Cleared by
A	Standard slant ( $0 < θ < 9 0^{\circ}$ )	the plain "plug in" case	Example 1
B	Complementary pair ( $θ$ and $9 0^{\circ} - θ$ )	same range, different path	Example 2
C	Degenerate: $θ = 0^{\circ}$	fired flat along ground	Example 3
D	Degenerate: $θ = 9 0^{\circ}$	straight up, $R = 0$	Example 3
E	Given $T$ or $H$ , find $u$ / $θ$	run the formula backwards	Example 4
F	Unequal heights (off a cliff)	ground formula banned, re-derive	Example 5
G	Real-world word problem	translate English → symbols	Example 6
H	Limiting behaviour ( $u \to$ big, $g \to$ small)	how answers scale	Example 7
I	Exam twist: two facts, hidden $θ$	combine $H$ and $R$	Example 8

Every cell A–I appears below at least once.

Example 1 — Cell A: the standard slant

Worked example Statement

A stone is launched at $u = 25 m/s$ , $θ = 3 7^{\circ}$ , with $g = 10 m/s^{2}$ . Find $T$ , $H$ , $R$ . Use $sin 3 7^{\circ} = 0.6$ , $cos 3 7^{\circ} = 0.8$ .

Forecast: before reading on, guess: will $H$ be more or less than $R$ ? (Hint: $R$ carries a $sin 2 θ$ , which is close to $sin 7 4^{\circ} \approx 0.96$ — nearly maximal — so $R$ will dwarf $H$ .)

Step 1 — Time of flight. Use $T = \frac{2 u sin θ}{g}$ . $T = \frac{2 ( 25 ) ( 0.6 )}{10} = \frac{30}{10} = 3 s$ Why this step? Vertical motion alone sets the clock; $sin θ$ picks out the upward slice of the launch speed.

Step 2 — Max height. Use $H = \frac{u ^{2} sin ^{2} θ}{2 g}$ . $H = \frac{( 25 ) ^{2} ( 0.6 ) ^{2}}{2 ( 10 )} = \frac{625 \cdot 0.36}{20} = \frac{225}{20} = 11.25 m$ Why this step? Squaring $sin θ$ AND squaring $u$ — height is "energy-like", it grows fast with speed.

Step 3 — Range. Use $R = \frac{u ^{2} sin 2 θ}{g}$ with $2 θ = 7 4^{\circ}$ , $sin 7 4^{\circ} \approx 0.9613$ . $R = \frac{625 ( 0.9613 )}{10} \approx 60.1 m$ Why this step? Range = steady forward speed $\times$ time aloft; the identity $2 sin θ cos θ = sin 2 θ$ folds both into one clean term.

Verify: dimensions — $T$ is $\frac{m/s}{m/s ^{2}} = s$ ✓; $H, R$ are $\frac{( m/s ) ^{2}}{m/s ^{2}} = m$ ✓. Cross-check $R$ the long way: $u cos θ \cdot T = 25 (0.8) (3) = 60 m$ , matching $60.1$ (rounding) ✓. And $R ≫ H$ as forecast ✓.

Example 2 — Cell B: the complementary twin

Worked example Statement

Fire the same stone ( $u = 25$ , $g = 10$ ) but at $θ = 5 3^{\circ}$ instead. Show its range equals Example 1's, yet its flight is taller and longer-lasting. ( $sin 5 3^{\circ} = 0.8, cos 5 3^{\circ} = 0.6$ .)

Forecast: $5 3^{\circ} = 9 0^{\circ} - 3 7^{\circ}$ . Guess: which of $T, H, R$ stays the same?

Step 1 — Range. $2 θ = 10 6^{\circ}$ , and $sin 10 6^{\circ} = sin (18 0^{\circ} - 10 6^{\circ}) = sin 7 4^{\circ} \approx 0.9613$ . $R = \frac{625 ( 0.9613 )}{10} \approx 60.1 m$ Why this step? $sin$ is mirror-symmetric about $9 0^{\circ}$ , so $2 θ$ and $18 0^{\circ} - 2 θ$ give the same range — that is why complementary angles are twins.

Step 2 — Height. $H = \frac{625 ( 0.8 ) ^{2}}{20} = \frac{625 \cdot 0.64}{20} = 20 m$ . Why this step? Now $sin θ$ is bigger (0.8 vs 0.6), so the upward launch slice is bigger — it climbs higher.

Step 3 — Time. $T = \frac{2 ( 25 ) ( 0.8 )}{10} = 4 s$ . Why this step? More upward speed ⇒ longer to rise and fall.

Verify: $R$ identical to Example 1 (60.1 m) ✓ — the twin. But $H$ rose from 11.25 m → 20 m and $T$ from 3 s → 4 s: same landing spot, a taller, slower arc. Sanity: $u cos θ \cdot T = 25 (0.6) (4) = 60$ m ✓.

Look at the two arcs below. The figure draws both throws on the same board: the chalk-blue curve is the $3 7^{\circ}$ flat-fast shot, the chalk-pink curve is the $5 3^{\circ}$ tall-slow shot. Trace each with your finger and notice they meet at the same pale-yellow landing dot on the ground — the range is genuinely identical — while the pink one peaks almost twice as high. That single picture is the complementary-twin idea.

Figure — Range, max height, time of flight — all derived

The blue arc reaches its peak of $11.25$ m early and comes down fast; the pink arc arcs up to $20$ m and lingers, yet both land at $60.1$ m. Same landing spot, very different journey — exactly what "same range, different path" means.

Example 3 — Cells C & D: the two degenerate throws

Worked example Statement

With $u = 25$ , $g = 10$ , evaluate everything at (i) $θ = 0^{\circ}$ (dead flat) and (ii) $θ = 9 0^{\circ}$ (straight up).

Forecast: which quantities become zero in each case? Picture the motion first.

Case (i), $θ = 0^{\circ}$ : $sin 0 = 0$ , $cos 0 = 1$ , $sin (2 \cdot 0) = sin 0 = 0$ . $T = \frac{2 u ( 0 )}{g} = 0, H = \frac{u ^{2} ( 0 )}{2 g} = 0, R = \frac{u ^{2} ( 0 )}{g} = 0$ Why this step? A ball launched perfectly horizontally from ground level has zero upward speed, so it is already "landed" — no air time, no height, no travel. (Launch it off a table and you get Example 5's cliff case.)

Case (ii), $θ = 9 0^{\circ}$ : $sin 9 0^{\circ} = 1$ , $cos 9 0^{\circ} = 0$ , $sin 18 0^{\circ} = 0$ . $T = \frac{2 u ( 1 )}{g} = \frac{2 ( 25 )}{10} = 5 s, H = \frac{u ^{2} ( 1 )}{2 g} = \frac{625}{20} = 31.25 m, R = \frac{u ^{2} ( 0 )}{g} = 0$ Why this step? Straight up means all speed goes into climbing (max height) and none forward (zero range). It comes straight back down where it left — $R = 0$ is correct, not a mistake.

Verify: $θ = 9 0^{\circ}$ gives the biggest possible $H$ (since $sin^{2} θ$ maxes at 1) yet zero range — the perfect illustration that height and range are traded off ✓. Units all consistent ✓.

Example 4 — Cell E: running the formula backwards

Worked example Statement

A projectile launched at $θ = 3 0^{\circ}$ stays in the air for exactly $T = 2 s$ ( $g = 10$ ). Find the launch speed $u$ , then its range.

Forecast: we're given the output $T$ and must recover the input $u$ . Guess whether $u$ will be larger or smaller than $T \times g = 20$ .

Step 1 — Invert time of flight. Start from $T = \frac{2 u sin θ}{g}$ and solve for $u$ : $u = \frac{g T}{2 s i n θ} = \frac{( 10 ) ( 2 )}{2 ( 0.5 )} = \frac{20}{1} = 20 m/s$ Why this step? Algebra is reversible; the same formula that predicts $T$ from $u$ lets us solve for $u$ when $T$ is the known.

Step 2 — Now range. $R = \frac{u ^{2} sin 2 θ}{g} = \frac{400 sin 6 0 ^{\circ}}{10} = \frac{400 ( 0.8660 )}{10} \approx 34.64 m$ . Why this step? Once $u$ is known the projectile is fully determined; every other quantity follows.

Verify: feed $u = 20$ forward: $T = \frac{2 ( 20 ) ( 0.5 )}{10} = 2$ s ✓ — matches the given time. This is the same ball as the parent note's Example 1, so $R \approx 34.6$ m ✓.

Example 5 — Cell F: off a cliff (formula forbidden!)

Worked example Statement

A ball is thrown from the top of a $20 m$ cliff at $u = 15 m/s$ , $θ = 5 3^{\circ}$ above horizontal ( $g = 10$ , $sin 5 3^{\circ} = 0.8$ , $cos 5 3^{\circ} = 0.6$ ). Find the total time to hit the ground below and the horizontal distance from the base.

Forecast: the boxed $R = \frac{u ^{2} s i n 2 θ}{g}$ assumes it lands at launch height. Here it lands lower. Guess: will the true time be more or less than the flat-ground $T$ ?

Step 1 — Set up the full vertical equation. Take up as positive, launch point as origin. Landing is $20$ m below, so displacement $y = - 20$ . $y = u sin θ t - \frac{1}{2} g t^{2} \Rightarrow - 20 = 15 (0.8) t - 5 t^{2} = 12 t - 5 t^{2}$ Why this step? The ground formula is banned because launch and landing heights differ; we must solve the raw quadratic with $y \neq = 0$ .

Step 2 — Solve the quadratic $5 t^{2} - 12 t - 20 = 0$ . $t = \frac{12 \pm 144 + 400}{10} = \frac{12 \pm 544}{10} = \frac{12 \pm 23.32}{10}$ Take the positive root: $t = \frac{35.32}{10} \approx 3.53 s$ . Why this step? Two roots appear (one negative "time before launch" that is unphysical); we keep the positive one — the real landing moment.

Step 3 — Horizontal distance. Horizontal speed is constant: $u_{x} = 15 (0.6) = 9 m/s$ . $x = u_{x} t = 9 (3.53) \approx 31.8 m$ Why this step? No horizontal force ⇒ uniform motion the whole flight, cliff or not.

Verify: flat-ground time would be $T = \frac{2 ( 15 ) ( 0.8 )}{10} = 2.4$ s; our cliff time $3.53$ s is longer — correct, since it falls an extra 20 m ✓. Discriminant $544 > 0$ so real landing exists ✓. Plug $t = 3.53$ back: $12 (3.53) - 5 (3.53)^{2} = 42.4 - 62.3 \approx - 19.9 \approx - 20$ ✓.

Example 6 — Cell G: the real-world word problem

Worked example Statement

A firefighter's hose ejects water at $u = 30 m/s$ . Aimed at $θ = 4 5^{\circ}$ , will the jet clear a wall $22 m$ away, and how high is the water when it reaches the wall? ( $g = 10$ , $sin 4 5^{\circ} = cos 4 5^{\circ} \approx 0.7071$ .)

Forecast: $4 5^{\circ}$ gives maximum range. Guess whether $22$ m is comfortably inside that reach.

Step 1 — Translate English to symbols. "Clears the wall" means the range must exceed $22$ m. Range at $4 5^{\circ}$ : $sin 9 0^{\circ} = 1$ . $R = \frac{u ^{2} s i n 9 0 ^{\circ}}{g} = \frac{900}{10} = 90 m$ Why this step? $R$ tells us where the water lands at ground level ( $90$ m ≫ $22$ m), so it certainly passes the wall's horizontal position.

Step 2 — Find the time to reach the wall. $u_{x} = 30 (0.7071) = 21.21 m/s$ . $t = \frac{x}{u _{x}} = \frac{22}{21.21} \approx 1.037 s$ Why this step? We need the height at $x = 22$ , so first get the time the water is at that horizontal spot.

Step 3 — Height at that time. $u_{y} = 21.21 m/s$ . $y = u_{y} t - \frac{1}{2} g t^{2} = 21.21 (1.037) - 5 (1.037)^{2} \approx 21.99 - 5.38 \approx 16.6 m$ Why this step? Vertical position uses the free-fall equation; that height tells us whether the jet is above the wall's top.

Verify: water is $\approx 16.6$ m high at the wall — clears any wall under that. Max height of the whole arc is $H = \frac{900 ( 0.5 )}{20} = 22.5$ m, and $16.6 < 22.5$ ✓ (it hasn't peaked yet at $x = 22$ , sensible since peak is at $x = R /2 = 45$ m). Units metres ✓.

Example 7 — Cell H: limiting behaviour (how answers scale)

Worked example Statement

Keep $θ = 4 5^{\circ}$ fixed. (a) If launch speed doubles from $u$ to $2 u$ , what happens to $R$ and $H$ ? (b) On the Moon, gravity is about $g /6$ . How does that change $R$ ?

Forecast: guess — does doubling speed double the range, or more than double it?

Step 1 — Speed scaling. Both $R$ and $H$ carry $u^{2}$ : $\frac{R ( 2 u )}{R ( u )} = \frac{( 2 u ) ^{2}}{u ^{2}} = 4, \frac{H ( 2 u )}{H ( u )} = 4$ Why this step? The formulas depend on $u^{2}$ , not $u$ — so any factor on speed gets squared in the answer. Double speed ⇒ quadruple range and height.

Step 2 — Gravity scaling. $R \propto \frac{1}{g}$ : $\frac{R _{moon}}{R _{earth}} = \frac{1/ ( g /6 )}{1/ g} = 6$ Why this step? Weaker gravity means longer time aloft ( $T \propto 1/ g$ ) and the range inherits that $1/ g$ — the same throw goes six times farther on the Moon.

Verify: concrete check with $u = 20$ , $θ = 4 5^{\circ}$ . Earth: $R = \frac{400 ( 1 )}{10} = 40$ m. Double speed to 40: $R = \frac{1600}{10} = 160$ m $= 4 \times 40$ ✓. Moon ( $g = 10/6 \approx 1.667$ ): $R = \frac{400}{1.667} \approx 240$ m $= 6 \times 40$ ✓.

Example 8 — Cell I: the exam twist (two facts, hidden angle)

Worked example Statement

A projectile has maximum height $H = 15 m$ and horizontal range $R = 40 m$ ( $g = 10$ ). Find its launch angle $θ$ . (You are NOT told $u$ .)

Forecast: two unknowns ( $u, θ$ ), two facts. Guess: will a ratio of $H$ to $R$ magically cancel $u$ ?

Step 1 — Write both formulas. $H = \frac{u ^{2} s i n ^{2} θ}{2 g}, R = \frac{u ^{2} s i n 2 θ}{g} = \frac{2 u ^{2} s i n θ c o s θ}{g}$ Why this step? Laying them side by side reveals the shared $u^{2}$ — a clue that dividing will erase it.

Step 2 — Take the ratio $H / R$ to kill $u^{2}$ and $g$ . $\frac{H}{R} = \frac{u ^{2} s i n ^{2} θ / ( 2 g )}{2 u ^{2} s i n θ c o s θ / g} = \frac{s i n ^{2} θ}{4 s i n θ c o s θ} = \frac{s i n θ}{4 c o s θ} = \frac{t a n θ}{4}$ Why this step? The unknown speed vanishes, leaving a single equation in $θ$ alone. The leftover $\frac{s i n θ}{c o s θ}$ is $tan θ$ — the natural tool because it packages the whole angle into one ratio (see Vectors — resolving into components).

Step 3 — Solve for $θ$ . $\frac{15}{40} = \frac{t a n θ}{4} \Rightarrow tan θ = \frac{4 \cdot 15}{40} = 1.5 \Rightarrow θ = arctan (1.5) \approx 56. 3^{\circ}$ Why this step? $arctan$ answers "which angle has this tangent?" — undoing the tan to recover the launch angle.

Verify: recover $u$ from $H$ : $u^{2} = \frac{2 g H}{s i n ^{2} θ} = \frac{2 ( 10 ) ( 15 )}{s i n ^{2} 56. 3 ^{\circ}} = \frac{300}{0.692} \approx 433.6$ , so $u \approx 20.8$ m/s. Feed into $R$ : $R = \frac{433.6 s i n ( 112. 6 ^{\circ} )}{10} = \frac{433.6 ( 0.923 )}{10} \approx 40.0$ m ✓ — matches the given range.

Recall

Complementary-angle twins

θ

and

9 0^{\circ} - θ

give equal range because

sin 2 θ = sin (18 0^{\circ} - 2 θ)

Why can't you use the range box off a cliff?

Launch and landing heights differ;

R = \frac{u ^{2} s i n 2 θ}{g}

assumes they are equal.

How do you cancel unknown

u

when given

H

and

R

Form

H / R = \frac{t a n θ}{4}

, then

θ = arctan (4 H / R)

Doubling launch speed multiplies range by?

Four (range

\propto u^{2}

Connections

Range, max height, time of flight — all derived (parent)
Projectile Motion — components & independence
Equations of Motion (1D kinematics)
Vectors — resolving into components
Free Fall & Acceleration due to gravity
Trigonometric identities — double angle
Energy method — max height via conservation