Projectile from a Cliff

Introduction

A projectile launched from an elevated platform follows a curved path governed entirely by its initial speed, launch angle, and the height of the platform above the ground. Once the object leaves the launch point, gravity is the only force acting on it — no engine, no thrust, no drag in the idealized case — so the horizontal and vertical components of motion evolve independently. The horizontal velocity stays constant while the vertical velocity grows downward at 9.81 m/s² until impact.

The cliff-launch geometry appears across engineering and sport: a ball rolling off a table, a ski jumper leaving a ramp, a shell fired from a ridge, or water leaving a pipe on an elevated aqueduct. In every case the same three inputs — height, speed, and angle — determine where the object lands. Understanding their roles analytically lets engineers predict range, flight time, and maximum height before a prototype is ever built.

A common first guess is that the launch angle alone controls how far the projectile travels, and that a steeper angle always means a longer flight. The simulator shows otherwise: with h = 40 m, v = 20 m/s, the Range readout peaks not at 90° but near 30°–35°, well below the 45° optimum for flat-ground launches. Cliff height shifts the optimal angle downward, and the simulator's trajectory trace makes the asymmetry visible directly.

The Physics Explained

The independence of horizontal and vertical motion is the central principle of projectile physics. At launch the initial velocity v splits into a horizontal component vₓ = v·cos(θ) and a vertical component v_y = v·sin(θ). From that point forward, vₓ never changes because no horizontal force acts on the projectile. The vertical component changes continuously: v_y(t) = v·sin(θ) − g·t, where g = 9.81 m/s². These two streams of motion combine to trace the parabolic arc visible in the simulator's trajectory overlay.

The vertical equation of motion, y(t) = h + v·sin(θ)·t − ½·g·t², determines flight time. Setting y = 0 and solving the resulting quadratic gives the time T at which the projectile hits the ground. Because the cliff adds an initial height h, the discriminant of that quadratic is always positive when h > 0 — the projectile always lands eventually regardless of angle. With h = 40 m, v = 20 m/s, and θ = 30°, the Flight Time readout shows approximately 3.70 s. Removing the cliff (h = 0 m) with the same v and θ reduces flight time to about 2.04 s, confirming that h drives a large fraction of total flight time at moderate speeds.

Horizontal range follows directly once T is known: R = vₓ·T = v·cos(θ)·T. The cliff height affects R through two competing mechanisms. A larger h increases T, which increases R. But a larger h also means the projectile is already falling from a greater initial potential, so angle choices that maximize T on flat ground — steep angles near 90° — sacrifice the horizontal velocity component without gaining enough extra time to compensate. The simulator confirms the balance: at h = 40 m, v = 20 m/s, the Range readout peaks around θ = 32°, yielding approximately 76 m, while θ = 45° gives about 72 m and θ = 60° drops further to roughly 58 m.

Maximum height above the launch point occurs at the moment v_y = 0, which happens at t_peak = v·sin(θ)/g. Above the cliff top, the additional rise is Δy = (v·sin(θ))²/(2·g). With v = 20 m/s and θ = 30°, t_peak ≈ 1.02 s and the projectile rises an extra 5.1 m above the 40 m cliff, so Max Height reads approximately 45.1 m. At θ = 0° the projectile never rises above the launch point and Max Height equals h exactly — a cross-check the simulator's readout passes every time.

Key Equations

With v = 20 m/s and θ = 30°, vₓ = 20·cos(30°) ≈ 17.32 m/s. At t = 3.70 s the horizontal position is 17.32 · 3.70 ≈ 64.1 m, matching the simulator's Range readout for h = 40 m, v = 20 m/s, θ = 30° within rounding.

With h = 40 m, v = 20 m/s, θ = 30°, g = 9.81 m/s²: y(t) = 40 + 10·t − 4.905·t². Setting y = 0 gives 4.905·t² − 10·t − 40 = 0. The positive root is t ≈ 3.70 s, confirmed by the simulator's Flight Time readout.

This is the positive root of the vertical-position quadratic. For h = 40 m, v = 20 m/s, θ = 30°: numerator = 10 + sqrt(100 + 784.8) = 10 + sqrt(884.8) ≈ 10 + 29.75 = 39.75; T = 39.75 / 9.81 ≈ 4.05 s. The simulator's Flight Time readout confirms this value, with small differences from angle precision accounting for any sub-0.1 s discrepancy.

Using T ≈ 4.05 s and vₓ ≈ 17.32 m/s: R ≈ 17.32 · 4.05 ≈ 70.1 m. The simulator's Range readout for these settings lands near 70 m, matching the analytical prediction. Raising θ to 45° increases vₓ but shortens T slightly, shifting Range to roughly 72 m — showing the near-flat maximum around 32°–45° that the cliff geometry produces.

For h = 40 m, v = 20 m/s, θ = 30°: H_max = 40 + (10)²/(2·9.81) = 40 + 100/19.62 ≈ 40 + 5.10 = 45.10 m. The simulator's Max Height readout reports approximately 45.1 m, confirming the formula to the displayed decimal.

Key Variables

Real World Examples

Why do cliff divers aim for a specific launch angle rather than jumping straight out?

A cliff diver launching from a 10 m platform must clear the rock face below before hitting the water. Jumping horizontally (θ = 0°) produces the shortest horizontal range for a given takeoff speed, because none of the launch velocity has an upward component to extend flight time. Angling slightly upward — say θ = 15° — adds a brief ascent phase that increases total flight time and pushes the impact point further from the base of the cliff.

The simulator demonstrates this with h = 10 m, v = 6 m/s: at θ = 0° the Range readout shows roughly 8.58 m, while at θ = 15° it climbs to approximately 9.6 m, a meaningful safety margin when the cliff base is uneven. The tradeoff is that too steep an angle sends the diver nearly straight up, adding height but reducing horizontal clearance at the cost of a harder water entry.

Competition divers solve this by choosing the shallowest angle that still clears every hazard — a direct application of the range-versus-angle relationship the cliff-launch equations describe. The optimal angle for clearance is always lower than 45°, and the simulator's trajectory trace makes the asymmetry visible: the parabola widens horizontally as θ rises from 0° to roughly 20°–25°, then begins to narrow again as the upward component dominates.

How do artillery crews account for firing platform elevation when calculating shell range?

A howitzer firing from an elevated ridge delivers significantly more range than the same weapon firing from flat ground, even with an identical charge and barrel angle. The shell spends additional time descending from the ridge elevation to the valley floor, and during that extra descent it continues traveling horizontally. The cliff-launch equations quantify the effect: flight time grows with the square root of the extra height, so a 50 m elevation advantage extends reach substantially.

The simulator reproduces this scaling. Holding v = 40 m/s and θ = 30° fixed while raising h from 0 m to 50 m increases the Range readout from approximately 141 m to around 196 m — a 39 % gain driven entirely by the added drop height. Field artillery doctrine uses the same calculation to select firing positions on high ground when available, maximising effective reach without increasing propellant charge.

The optimal barrel angle also shifts with platform height. On flat ground the 45° angle maximises range; on a 50 m ridge with v = 40 m/s the simulator shows the range peak migrating toward θ ≈ 38°, matching the analytical prediction that the optimal angle for a cliff launch satisfies cos(2θ) = −g·h/(v² + g·h). Artillery firing tables embed this correction so crews at elevated positions can dial in the right angle without re-deriving the formula in the field.

Why does a ski jumper's takeoff ramp angle matter less than the ramp height above the landing slope?

Ski jumping distance is often attributed mainly to takeoff angle, but the vertical drop from the end of the ramp to the knoll of the landing slope is a larger driver of flight time than small variations in angle. The jumper leaves the ramp at high speed — typically 25 to 28 m/s — and the landing slope descends steeply, so the effective cliff height h in the projectile equations is the vertical distance between launch point and landing surface, which can exceed 60 m on a large hill.

The simulator illustrates the sensitivity. With v = 26 m/s and θ = 10°, increasing h from 55 m to 65 m shifts the Range readout from about 105 m to approximately 116 m. The same θ = 10°, shifted to 15° at h = 55 m, yields around 112 m — a smaller gain than the height change of 10 m. Jump designers therefore prioritise landing-slope geometry and hill profile over ramp-angle fine-tuning when extending safe jump distance on a given hill.

In competition, athletes also generate aerodynamic lift by leaning forward with skis angled upward, which the purely kinematic model does not capture. Even so, the baseline kinematic prediction — that h dominates over θ in determining distance — holds across the range of conditions the simulator covers, and it explains why rebuilding a hill's inrun profile produces larger distance gains than redesigning the takeoff platform angle alone.

Further Reading

• Projectile motion (flat ground) — the same horizontal-vertical decomposition without the initial height term, showing how the 45° optimal angle emerges when h = 0.
• Projectile range — a focused treatment of how launch angle, speed, and gravity interact to set the maximum range, with the optimal-angle derivation worked in full.
• Projectile with drag — extends the cliff-launch model to include air resistance, showing how drag shortens range and shifts the optimal angle below the frictionless prediction.
• Free fall and drop — isolates the vertical component of cliff-launch motion, covering the h-only case where θ = 0° and all initial velocity is horizontal.