What is the formula for the range of a projectile?

For a ground-level launch, the range is R = v²·sin(2θ)/g, where v is the initial speed, θ is the launch angle, and g is gravitational acceleration (9.81 m/s² on Earth).

Why does air resistance reduce range and break parabolic symmetry?

Air drag opposes velocity (Fd = ½CdρAv²), removing kinetic energy continuously. This reduces speed, shortens range, lowers max height, and makes the descending arc steeper than the ascending one, breaking the perfect parabola.

Why is the speed at the highest point equal to v·cosθ?

At the peak, the vertical velocity component is zero (the projectile momentarily has no vertical motion). Only the horizontal component v·cosθ remains, so the speed at the highest point is exactly v·cosθ.

Mechanics 28 April 2026

Projectile Motion Simulator — Range, Height & Time Calculator

Q: How does launch height affect range?

Launching from a height h increases the time of flight, so the projectile travels farther horizontally. The range is R = (v·cosθ/g)(v·sinθ + √(v²sin²θ + 2gh)).

Whether you are studying physics at secondary school, an engineering undergraduate, or a lecturer preparing demonstrations, this guide walks you through every formula that governs a flying object — then lets you verify them live in the Projectile Motion Simulator. We cover range, max height, time of flight, velocity decomposition, the effect of launch height, and real-world drag.

What Is Projectile Motion?

Projectile motion describes the curved path of any object that is launched into the air and then moves only under the influence of gravity (and possibly air resistance). Once the object leaves your hand — or the cannon, or the cliff — no engine or force acts horizontally. Gravity pulls it steadily downward, creating a parabolic arc.

The key insight, first stated clearly by Galileo, is that horizontal and vertical motions are completely independent. A bullet fired horizontally and one dropped from the same height hit the ground at exactly the same moment, because both experience the same gravitational acceleration in the vertical direction.

The Core Equations — Range, Height, Time

For a projectile launched from ground level with initial speed v at angle θ, the equations below follow directly from integrating Newton's second law:

Horizontal range

\[R = \dfrac{v^2 \sin 2\theta}{g}\]

Maximum height

\[H = \dfrac{v^2 \sin^2\!\theta}{2g}\]

Time of flight

\[T = \dfrac{2v \sin\theta}{g}\]

Parabolic trajectory

\[y = x\tan\theta - \dfrac{gx^2}{2v^2\cos^2\!\theta}\]

Notice that the range formula uses sin 2θ: this term is largest when 2θ = 90°, i.e. θ = 45°. Complementary angles like 30° and 60° give identical ranges because sin(60°) = sin(120°).

Worked Example: Artillery Shell at 45°

The simulator's Artillery preset fires at v = 80 m/s, θ = 45°, from ground level. Let us verify the readouts by hand:

Range: R = 80² × sin(90°)/9.81 = 6400/9.81 = 652 m
Max height: H = 6400 × sin²(45°)/(2 × 9.81) = 6400 × 0.5/19.62 = 163 m
Time of flight: T = 2 × 80 × sin(45°)/9.81 = 160 × 0.707/9.81 = 11.5 s
Impact speed: 80 m/s (equal to launch speed for ground-level launch, by energy conservation)

Load the Artillery preset in the simulator and confirm all four readouts match.

Velocity Decomposition and Independence of Motion

The initial velocity vector splits into two independent components:

\[V_x = v\cos\theta \qquad V_{y_0} = v\sin\theta\]

The horizontal component V_x never changes (no horizontal force). The vertical component decreases by g every second:

\[V_y(t) = v\sin\theta - g\,t\]

At the highest point V_y = 0, so the speed there is simply V_x = v·cos θ. For our 80 m/s, 45° artillery shell that is 80 × cos(45°) = 56.6 m/s.

The simulator draws the velocity vector live at every time step — set a low launch speed such as 10 m/s so you can watch V_y shrink to zero at the apex and then grow downward.

Effect of Launch Height

When the launch height h > 0, the projectile has extra time to travel horizontally while falling to ground level. The range formula becomes:

\[R = \dfrac{v\cos\theta}{g}\!\left(v\sin\theta + \sqrt{v^2\sin^2\!\theta + 2gh}\right)\]

The Basketball Shot preset (v = 8 m/s, θ = 52°, h = 2 m) captures a real shooting scenario. The horizontal component is Vx = 8 × cos(52°) = 4.93 m/s and the ball travels roughly 7.6 m before landing — close to the typical three-point-line distance when accounting for the basket height.

Projectile Motion Simulator showing a basketball shot at v=8 m/s, 52°, from h=2 m with a realistic arc trajectory

As launch height increases, the optimal angle (the one maximising range) decreases below 45°. This explains why football throw-ins from a height use a flatter angle than a field kick from ground level.

Air Resistance and Real-World Scenarios

Activate the Air Resistance toggle in the simulator to apply a quadratic drag model:

\[F_d = \tfrac{1}{2}\,C_d\,\rho\,A\,v^2\]

where C_d = 0.47 (sphere), ρ = 1.225 kg/m³ (air density), and A = π r² (cross-sectional area of a 5 cm radius ball). The drag force continuously removes energy and acts opposite to the velocity vector, so the descending arc becomes steeper than the ascending one — the parabola symmetry is broken. The optimal angle drops from 45° to roughly 30–38° depending on projectile and speed.

The Golf Drive preset (v = 70 m/s, θ = 12°) illustrates this perfectly: a low, flat angle is used precisely because backspin and drag interact to extend carry rather than height.

Using the Simulator — Four Preset Scenarios

Open the Projectile Motion Simulator and explore the four presets to build physical intuition quickly:

Preset	v (m/s)	θ (°)	h (m)	Key lesson
Optimal 45°	30	45	0	Maximum range at ground level
Artillery	80	45	0	Scale of range with high speed
Basketball	8	52	2	Real trajectory with launch height
Golf Drive	70	12	0	Flat angle, drag-modified path

After loading each preset, switch on Air Resistance and compare how the range and trajectory change. The simulator's ghost trails let you overlay multiple launches for direct comparison.

Key Takeaways

The range formula \(R = v^2\sin 2\theta / g\) shows that 45° is optimal on flat ground.
Complementary angles (e.g. 30° and 60°) give equal ranges but different heights.
Launch height increases time of flight, extending range beyond the ground-level formula.
At the peak, the speed equals the horizontal component \(v\cos\theta\); vertical speed is zero.
Air drag reduces range, lowers max height, and breaks parabolic symmetry.
The optimal launch angle with drag is less than 45° — typically 30–38° for most sports balls.

Frequently Asked Questions

What angle maximises the range of a projectile?

On flat ground with no air resistance, 45° maximises range because sin(2θ) reaches its maximum of 1 at θ = 45°. With drag or from a height, the optimal angle is below 45°.

What is the formula for range?

For a ground-level launch: R = v²·sin(2θ)/g. With launch height h: R = (v·cosθ/g)(v·sinθ + √(v²sin²θ + 2gh)).

How does launch height affect range?

Greater launch height means longer time of flight, so more horizontal distance is covered. The optimal angle also decreases below 45° as h increases.

Why does air resistance break the parabolic symmetry?

Quadratic drag (Fd = ½CdρAv²) continuously removes energy. The projectile moves slower on the way down, making the descending arc steeper than the ascending one.

What is the speed at the highest point?

At the apex, vertical velocity is zero. Only the horizontal component remains: speed = v·cosθ. For v = 80 m/s, θ = 45°, that is 80 × cos(45°) ≈ 56.6 m/s.