Projectile motion — horizontal - vertical independence, full derivation: Worked examples

This page is a drill. The parent note built the machinery; here we throw every kind of ball at it. The plan: first map out all the case classes a projectile problem can be, then solve one example per case so you never meet a scenario you haven't seen.

The scenario matrix

Every projectile problem is one of these cells. Each row is a way the setup can differ; the examples that follow are tagged with the cell they cover.

Cell	What makes it special	Danger it hides	Example
A Level launch, generic $θ$	launch height = landing height	none — plug into $T, H, R$	Ex 1
B $θ = 0$ (horizontal throw)	$u_{y} = 0$ from the start	can't use $R = u^{2} sin 2 θ / g$	Ex 2
C $θ = 9 0^{\circ}$ (straight up)	$u_{x} = 0$ , range is zero	it's really 1-D	Ex 3
D Landing below launch (cliff)	$y_{final} < 0$	naive $T, R$ formulas fail	Ex 4
E Landing above launch (onto a wall/roof)	$y_{final} > 0$	two candidate times	Ex 5
F Complementary angles	$θ$ and $9 0^{\circ} - θ$	same range, different $T, H$	Ex 6
G Velocity/direction during flight	need $v_{x}, v_{y}$ at time $t$	sign of $v_{y}$ flips at the top	Ex 7
H Real-world word problem	hidden numbers, unit traps	reading, not physics	Ex 8
I Exam twist — solve for $θ$	given $R$ , find angle	$sin$ gives two answers	Ex 9
J Degenerate / limit check	$u = 0$ , $g \to$ large, $θ \to 0$	formulas must stay sane	Ex 10

Cell A — Level launch, generic angle

Worked example Example 1 — the vanilla case

A ball leaves the ground at $u = 20 m/s$ , $θ = 3 0^{\circ}$ , $g = 10 m/s^{2}$ . Find flight time $T$ , max height $H$ , range $R$ , and the speed at landing.

Forecast: guess before computing — will the landing speed be more, less, or equal to $20 m/s$ ?

Split the launch velocity. $u_{x} = 20 cos 3 0^{\circ} = 20 (0.866) = 17.32 m/s$ , $u_{y} = 20 sin 3 0^{\circ} = 20 (0.5) = 10 m/s$ . Why this step? The two axes are independent; we must feed each its own starting speed.
Flight time $T = \frac{2 u _{y}}{g} = \frac{2 ( 10 )}{10} = 2 s$ . Why this step? The ball returns to $y = 0$ when vertical motion completes its up-and-down; only $u_{y}$ decides this.
Max height $H = \frac{u _{y}^{2}}{2 g} = \frac{100}{20} = 5 m$ . Why this step? At the top $v_{y} = 0$ ; energy/kinematics of the vertical piece alone give $H$ .
Range $R = u_{x} T = 17.32 (2) = 34.64 m$ . Why this step? Horizontal speed is constant, so distance = speed × total time.
Landing speed. By symmetry $v_{x} = 17.32$ , $v_{y} = - 10$ (same magnitude, now downward). Speed $= 17.3 2^{2} + 1 0^{2} = 300 + 100 = 20 m/s$ .

Verify: landing speed equals launch speed $20 m/s$ — correct, because on level ground gravity gives back exactly the kinetic energy it took. Forecast answer: equal. Units: m/s ✓.

Cell B — Horizontal throw ( $θ = 0$ )

Worked example Example 2 — drop vs shoot

A stone is thrown horizontally at $u = 15 m/s$ from a $20 m$ cliff. When and where does it land? Compare to a stone simply dropped.

Forecast: does the thrown stone hit the ground before the dropped one, because it "has more going on"?

Resolve. $θ = 0 \Rightarrow u_{x} = 15$ , $u_{y} = 0$ . Why this step? A flat throw pours all its speed into horizontal; vertical starts from rest.
Vertical fall. Take down as positive here: $20 = \frac{1}{2} (10) t^{2} \Rightarrow t^{2} = 4 \Rightarrow t = 2 s$ . Why this step? Vertically the stone is a pure free-fall object; horizontal speed is irrelevant to falling.
Horizontal distance. $x = u_{x} t = 15 (2) = 30 m$ . Why this step? Constant horizontal velocity across the shared clock $t$ .
Dropped stone. $20 = \frac{1}{2} (10) t^{2} \Rightarrow t = 2 s$ — identical. Why this step? The dropped stone also has $u_{y} = 0$ and falls the same $20 m$ under the same $g$ ; its vertical equation is byte-for-byte the same as the thrown stone's, so it must give the same $t$ . This is the visible proof of independence.

Verify: both take $2 s$ ; the horizontal throw does not change fall time — this is the whole point of independence. Forecast answer: no, they hit together. Units ✓.

Cell C — Straight up ( $θ = 9 0^{\circ}$ )

Cell D — Landing below launch (cliff)

Worked example Example 4 — launch off a cliff at an angle

From the top of a $25 m$ cliff, a ball is thrown at $u = 20 m/s$ , $θ = 3 0^{\circ}$ above horizontal. How long is it in the air, and how far from the cliff base does it land?

Forecast: will the flight time be more or less than the level-ground $T = 2 s$ from Example 1?

The figure below shows the setup: the ball launches from the cliff top, arcs up past launch level, then keeps falling the extra $25 m$ to the base. Notice the red arc extends below the dashed launch line — that extra fall is exactly why the naive $T$ formula is wrong here.

Resolve. $u_{x} = 20 cos 3 0^{\circ} = 17.32$ , $u_{y} = 20 sin 3 0^{\circ} = 10$ .
Set the vertical equation to the true landing height. The base of the cliff is $25 m$ below launch, so $y = - 25$ : $- 25 = 10 t - \frac{1}{2} (10) t^{2} \Rightarrow 5 t^{2} - 10 t - 25 = 0 \Rightarrow t^{2} - 2 t - 5 = 0.$ Why this step? The $T = 2 u sin θ / g$ formula assumes landing at launch height — here it doesn't, so we go back to the raw $y (t)$ equation.
Solve the quadratic. $t = \frac{2 \pm 4 + 20}{2} = 1 \pm 6$ . Positive root: $t = 1 + 2.449 = 3.449 s$ . Why this step? Only the positive time is physical (the negative root is the parabola's "past").
Horizontal distance. $x = u_{x} t = 17.32 (3.449) = 59.73 m$ . Why this step? Now that the vertical motion has told us when it lands, the horizontal position at that instant is just constant speed × that time — the shared clock $t$ links the two axes.

Verify: the ball spends longer aloft ( $3.45 s > 2 s$ ) because it keeps falling past the launch level — matches the figure's extended red arc. Forecast answer: more. Units ✓.

Cell E — Landing above launch (onto a wall)

Worked example Example 5 — clearing a wall (two candidate times)

A ball is launched from the ground at $u = 25 m/s$ , $θ = 5 3^{\circ}$ ( $sin 5 3^{\circ} = 0.8$ , $cos 5 3^{\circ} = 0.6$ ). A wall of height $15 m$ stands somewhere ahead. At what times is the ball at height $15 m$ ?

Forecast: how many distinct times can a projectile be at a given height below its peak?

Resolve. $u_{x} = 25 (0.6) = 15$ , $u_{y} = 25 (0.8) = 20$ .
Set $y = 15$ . $15 = 20 t - 5 t^{2} \Rightarrow 5 t^{2} - 20 t + 15 = 0 \Rightarrow t^{2} - 4 t + 3 = 0$ . Why this step? We want every instant the ball is at that height, so solve the full quadratic — not just one root.
Factor. $(t - 1) (t - 3) = 0 \Rightarrow t = 1 s$ (going up) and $t = 3 s$ (coming down). Why this step? A projectile passes each sub-peak height twice — once rising, once falling. The two roots are both real and physical.
Peak check. Max height $H = \frac{u _{y}^{2}}{2 g} = \frac{2 0 ^{2}}{2 ( 10 )} = 20 m > 15 m$ , so $15 m$ really is reachable. Why this step? If the wall were taller than $H$ the quadratic would have no real roots (the ball never gets there); computing $H$ confirms our two times are genuine, not algebraic ghosts.

Verify: the two times are symmetric about $t_{top} = \frac{u _{y}}{g} = 2 s$ ; indeed $\frac{1 + 3}{2} = 2$ ✓. Forecast answer: two (if below peak). Units ✓.

Cell F — Complementary angles

Worked example Example 6 — same range, different everything else

A cannon fires at $u = 40 m/s$ , once at $θ = 3 0^{\circ}$ and once at $θ = 6 0^{\circ}$ . Compare $R$ , $T$ , and $H$ .

Forecast: which of the three quantities are equal for the two shots?

The figure overlays both arcs. Notice they meet again at the same landing point on the ground (equal range, yellow line) even though the $6 0^{\circ}$ arc towers far above the $3 0^{\circ}$ one — a striking picture of "same $R$ , different $H$ ".

Ranges. $R = \frac{u ^{2} sin 2 θ}{g}$ . For $3 0^{\circ}$ : $sin 6 0^{\circ} = 0.866$ , $R = \frac{1600 ( 0.866 )}{10} = 138.6 m$ . For $6 0^{\circ}$ : $sin 12 0^{\circ} = 0.866$ , same $R = 138.6 m$ . Why this step? $sin 2 θ = sin (18 0^{\circ} - 2 θ)$ , so complementary angles share $sin 2 θ$ — equal range.
Times. $T = \frac{2 u sin θ}{g}$ . $3 0^{\circ}$ : $T = \frac{2 ( 40 ) ( 0.5 )}{10} = 4 s$ . $6 0^{\circ}$ : $T = \frac{2 ( 40 ) ( 0.866 )}{10} = 6.93 s$ . Why this step? $T \propto sin θ$ , and $sin 6 0^{\circ} > sin 3 0^{\circ}$ — the steeper shot lingers.
Heights. $H = \frac{u ^{2} sin ^{2} θ}{2 g}$ . $3 0^{\circ}$ : $\frac{1600 ( 0.25 )}{20} = 20 m$ . $6 0^{\circ}$ : $\frac{1600 ( 0.75 )}{20} = 60 m$ . Why this step? $H = \frac{u _{y}^{2}}{2 g}$ with $u_{y} = u sin θ$ gives $H = \frac{u ^{2} sin ^{2} θ}{2 g}$ ; since $H$ depends on $sin^{2} θ$ , the steeper $6 0^{\circ}$ shot (bigger $sin θ$ ) climbs far higher.

Verify: ranges equal (138.6 m each) ✓; the $6 0^{\circ}$ shot flies 3× higher and stays up longer — see the tall vs flat arcs. Forecast answer: only range is equal. Units ✓.

Cell G — Velocity and direction during flight

Worked example Example 7 — speed and direction mid-flight

For the Example 1 ball ( $u_{x} = 17.32$ , $u_{y} = 10$ , $g = 10$ ), find the velocity vector at $t = 1.5 s$ : its components, speed, and the angle it makes with the horizontal.

Forecast: at $t = 1.5 s$ (after the $t = 1 s$ peak) is the ball moving up or down?

Horizontal velocity. $v_{x} = u_{x} = 17.32 m/s$ (constant, unchanged). Why this step? $a_{x} = 0$ ; horizontal speed never varies — the classic trap.
Vertical velocity. $v_{y} = u_{y} - g t = 10 - 10 (1.5) = - 5 m/s$ . Why this step? Negative means downward — the ball is past its peak (peak was at $t = 1 s$ ).
Speed. $∣ v ∣ = v_{x}^{2} + v_{y}^{2} = 17.3 2^{2} + 5^{2} = 300 + 25 = 325 = 18.03 m/s$ . Why this step? The velocity is a vector; its two perpendicular components form the legs of a right triangle, so the actual speed is the hypotenuse (Pythagoras) — you cannot just add $v_{x}$ and $v_{y}$ .
Direction below horizontal. $α = arctan \frac{∣ v _{y} ∣}{v _{x}} = arctan \frac{5}{17.32} = 16. 1^{\circ}$ below horizontal. Why this step? The velocity vector's angle is opposite/adjacent on the right triangle its components form; $arctan$ asks "which angle has this tangent?" See Vectors — Resolving into Components.

Verify: $v_{y} < 0$ confirms descent; speed $18.03 < 20 m/s$ (launch) because it isn't back to launch height yet. Forecast answer: down. Units ✓.

Cell H — Real-world word problem

Worked example Example 8 — the water fountain

A garden fountain shoots water at $u = 14 m/s$ , $θ = 4 5^{\circ}$ . The gardener wants a circular pool that just catches all the water. What minimum pool radius (measured from the nozzle) is needed? ( $g = 9.8 m/s^{2}$ .)

Forecast: does the largest splash land at $4 5^{\circ}$ or would a different angle reach farther?

Identify the hidden ask. "Just catches all the water" = the range of the jet. Why this step? Word problems bury the physics quantity in plain language; range is what "how far it lands" means.
Use the range formula. $R = \frac{u ^{2} sin 2 θ}{g} = \frac{1 4 ^{2} sin 9 0 ^{\circ}}{9.8} = \frac{196 ( 1 )}{9.8} = 20 m$ . Why this step? Level ground (nozzle and pool at same height) ⇒ the standard $R$ applies, and $θ = 4 5^{\circ}$ makes $sin 2 θ = 1$ .
Answer. Radius $= 20 m$ .

Verify: $4 5^{\circ}$ is the max-range angle, so this is also the worst case radius — smaller angles land nearer. Units: $\frac{( m/s ) ^{2}}{m/s ^{2}} = m$ ✓. Forecast answer: $4 5^{\circ}$ is farthest.

Cell I — Exam twist: solve for the angle

Worked example Example 9 — given the range, find the launch angle

A ball thrown at $u = 30 m/s$ lands $45 m$ away on level ground ( $g = 10$ ). What launch angle(s) achieve this?

Forecast: how many launch angles can hit the same spot?

Set up the range equation for $θ$ . $45 = \frac{3 0 ^{2} sin 2 θ}{10} = 90 sin 2 θ \Rightarrow sin 2 θ = 0.5$ . Why this step? We now know $R$ and $u$ ; the unknown is inside $sin 2 θ$ , so isolate it.
Invert the sine. $2 θ = arcsin (0.5) = 3 0^{\circ}$ or $2 θ = 18 0^{\circ} - 3 0^{\circ} = 15 0^{\circ}$ . Why this step? $sin$ takes the same value at an angle and its supplement — that's the source of the two solutions.
Solve. $θ = 1 5^{\circ}$ or $θ = 7 5^{\circ}$ . Why this step? These are complementary ( $1 5^{\circ} + 7 5^{\circ} = 9 0^{\circ}$ ) — exactly the Cell F pattern.

Verify: plug $θ = 1 5^{\circ}$ : $sin 3 0^{\circ} = 0.5$ , $R = 90 (0.5) = 45 m$ ✓; same for $7 5^{\circ}$ . Forecast answer: two (a low and a high shot). Units ✓.

Cell J — Degenerate & limiting checks

Worked example Example 10 — sanity limits (does the maths stay honest?)

Test the formulas at three extremes and confirm they behave.

Forecast: should $R$ grow or shrink if we crank up $g$ ?

Zero speed, $u = 0$ . $R = \frac{0 \cdot sin 2 θ}{g} = 0$ , $H = \frac{0}{2 g} = 0$ , $T = \frac{0}{g} = 0$ . Why this step? No launch ⇒ no motion; a well-behaved formula must return exactly zero for every quantity, and each does.
Angle $\to 0$ . As $θ \to 0$ , $sin θ \to 0 \Rightarrow T = \frac{2 u sin θ}{g} \to 0$ and $H = \frac{u ^{2} sin ^{2} θ}{2 g} \to 0$ ; and $R = \frac{u ^{2} sin 2 θ}{g} \to 0$ too. On truly flat ground a horizontal throw from ground level has nowhere to rise, so it "lands instantly." Why this step? This checks the boundary between Cell A (angled level launch) and Cell B (horizontal throw): on flat ground the horizontal case degenerates, which is exactly why Cell B needs a cliff to be interesting.
Stronger gravity, $g \to$ large. With $u = 20$ , $θ = 4 5^{\circ}$ : $g = 10 \Rightarrow R = \frac{400 ( 1 )}{10} = 40 m$ ; $g = 20 \Rightarrow R = \frac{400}{20} = 20 m$ . Why this step? Doubling $g$ halves the range — matches the intuition that a heavier-pulling planet yanks the ball down sooner, cutting the flight short.

Verify: all degenerate outputs are finite and non-negative; $R$ shrinks as $g$ grows ( $40 \to 20 m$ ), and every $u = 0$ or $θ \to 0$ output collapses to $0$ as physics demands. Forecast answer: shrink. Units ✓.

Active recall

Recall Which cell am I in?

Ball thrown flat off a rooftop → ::: Cell D (landing below launch — use full $y (t)$ , not $T = 2 u sin θ / g$ ).
"At what two times is it at height $h$ ?" → ::: Cell E (quadratic, two roots symmetric about the peak time).
Given $R$ and $u$ , find $θ$ → ::: Cell I ( $sin 2 θ$ gives two complementary angles).
Ball fired straight up → ::: Cell C (range zero, really 1-D).