📐 Mechanics  ·  15 January 2026

Projectile Motion

Projectile motion is one of the first problems you encounter in classical mechanics — and one of the most elegant. An object is launched into the air with some initial speed v0v_0 at an angle θ\theta above the horizontal. After that moment, the only force acting on it is gravity (we ignore air resistance).

Because gravity acts only downward, the motion separates neatly into two independent components:

ComponentAccelerationVelocityPosition
Horizontal (x)00v0x=v0cosθv_{0x} = v_0\cos\theta (constant)x=v0cosθ  tx = v_0 \cos\theta\; t
Vertical (y)g-gv0y=v0sinθgtv_{0y} = v_0\sin\theta - g\,ty=v0sinθ  t12gt2y = v_0 \sin\theta\; t - \tfrac{1}{2}g\,t^2

Combining these gives the parabolic trajectory:

x(t)=v0cosθ  ty(t)=v0sinθ  t12gt2x(t) = v_0 \cos\theta \; t \qquad y(t) = v_0 \sin\theta \; t - \tfrac{1}{2}g\,t^2

Key Derived Quantities

From these equations we can derive three important results (assuming level ground, y0=0y_0 = 0):

  • Time of flight: T=2v0sinθg\displaystyle T = \frac{2\,v_0 \sin\theta}{g}
  • Maximum height: H=v02sin2θ2g\displaystyle H = \frac{v_0^2 \sin^2\theta}{2g}
  • Range: R=v02sin2θg\displaystyle R = \frac{v_0^2 \sin 2\theta}{g}

Notice that the range depends on sin2θ\sin 2\theta, which is maximised when 2θ=90°2\theta = 90°, i.e. θ=45°\theta = 45°. So for any given launch speed, a 45° angle gives the longest range (in the absence of air resistance).

Symmetry insight: Launch angles that add up to 90° (e.g. 30° and 60°) give the same range but very different trajectories. Try it below!


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50 m/s
45 °
9.81 m/s²

Time of Flight: 7.21 s

Max Height: 63.71 m

Range: 254.84 m


What to Observe

  • Angle vs Range: Set velocity to 50 m/s. Sweep the angle from 10° to 80°. Notice the range peaks at 45°, and complementary angles (e.g. 30° & 60°) give the same range but different peak heights.
  • Gravity's role: Lower gravity (think Moon at ~1.62 m/s²) dramatically increases both range and height. On Jupiter (~24.8 m/s²), the same throw barely gets off the ground.
  • Speed matters quadratically: Doubling v0v_0 quadruples the range (Rv02R \propto v_0^2). That's why a small increase in launch speed makes a huge difference.

Real-World Applications

  • Sports: Every ball sport involves projectile motion — from football goal-kicks to basketball free throws and cricket sixes.
  • Artillery & Rocketry: Ballistic trajectories were the original motivation for studying this problem (Galileo, 1638).
  • Space launches: Orbital mechanics begins where projectile motion meets the curvature of the Earth.

Limitations of This Model

This simulation assumes no air resistance. In reality, drag force (Fd=12CdρAv2F_d = \tfrac{1}{2} C_d \rho A v^2) slows the projectile, reduces the range, and makes the trajectory asymmetric — the descending arc is steeper than the ascending one.


Worked Examples

Worked Example

Example 1 — Football kicked at 30°

A footballer kicks a ball with an initial speed of 25 m/s at an angle of 30° above the ground. Find the range and maximum height (ignore air resistance, g = 9.81 m/s²).

Given: v₀ = 25 m/s, θ = 30°, g = 9.81 m/s²

Time of flight: T=2v0sinθg=2×25×sin30°9.81=2×25×0.59.812.55 sT = \frac{2 v_0 \sin\theta}{g} = \frac{2 \times 25 \times \sin 30°}{9.81} = \frac{2 \times 25 \times 0.5}{9.81} \approx 2.55 \text{ s}

Range: R=v02sin2θg=625×sin60°9.81625×0.8669.8155.1 mR = \frac{v_0^2 \sin 2\theta}{g} = \frac{625 \times \sin 60°}{9.81} \approx \frac{625 \times 0.866}{9.81} \approx 55.1 \text{ m}

Maximum height: H=v02sin2θ2g=625×0.252×9.817.97 mH = \frac{v_0^2 \sin^2\theta}{2g} = \frac{625 \times 0.25}{2 \times 9.81} \approx 7.97 \text{ m}

Worked Example

Example 2 — Ball thrown horizontally from a cliff

A ball is thrown horizontally at 15 m/s from the edge of a cliff 45 m high. How far from the base of the cliff does it land?

Given: v₀ = 15 m/s, θ = 0°, height h = 45 m, g = 9.81 m/s²

Time to fall (using vertical motion: h = ½gt²): t=2hg=2×459.819.173.03 st = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 45}{9.81}} \approx \sqrt{9.17} \approx 3.03 \text{ s}

Horizontal range: R=v0×t=15×3.0345.4 mR = v_0 \times t = 15 \times 3.03 \approx 45.4 \text{ m}

The ball lands approximately 45.4 m from the base of the cliff.


Launching from a Height

So far we assumed the ball starts at ground level (y0=0y_0 = 0). What if it's launched from a cliff, a platform, or a moving vehicle at height hh?

The horizontal equation is unchanged. Only the vertical equation gains an initial offset:

x(t)=v0cosθ  ty(t)=h+v0sinθ  t12gt2x(t) = v_0 \cos\theta \; t \qquad y(t) = h + v_0 \sin\theta \; t - \tfrac{1}{2}g\,t^2

Setting y=0y = 0 and solving the resulting quadratic gives the time of flight:

T=v0sinθ+v02sin2θ+2ghgT = \frac{v_0 \sin\theta + \sqrt{v_0^2 \sin^2\theta + 2gh}}{g}

The range is then R=v0cosθTR = v_0 \cos\theta \cdot T. Notice that with h>0h > 0, a launch angle slightly below 45° now maximises range — the ball has extra height to trade for forward distance.


Air Resistance and Drag

Real projectiles push through air. The drag force opposes the velocity vector and grows with the square of speed:

Fdrag=βmv2v^\mathbf{F}_\text{drag} = -\beta m v^2 \hat{\mathbf{v}}

where β\beta (m⁻¹) is the drag coefficient per unit mass and v=vv = |\mathbf{v}|. Splitting into components:

ax=βvvxay=gβvvya_x = -\beta\, v\, v_x \qquad a_y = -g - \beta\, v\, v_y

These coupled equations have no closed-form solution. Instead we use Euler's method — divide time into small steps Δt\Delta t and update iteratively:

vxvx+axΔtvyvy+ayΔtv_x \leftarrow v_x + a_x\,\Delta t \qquad v_y \leftarrow v_y + a_y\,\Delta t xx+vxΔtyy+vyΔtx \leftarrow x + v_x\,\Delta t \qquad y \leftarrow y + v_y\,\Delta t

Repeat until y<0y < 0. The simulations below use Δt=0.01 s\Delta t = 0.01\text{ s}, which is accurate enough for visualisation.

Observable effects of drag: shorter range, lower peak height, asymmetric trajectory (steeper descent than ascent), and an optimal launch angle that drops below 45°. Toggle air resistance in the panels below to see these effects directly.

Note: A full treatment of Euler integration — accuracy, step-size error, and higher-order alternatives — will be covered in a dedicated article.


Angle Sweep

See how launch angle shapes the trajectory — and why 45° gives the longest range (and how height and drag shift that optimum).

45 °
50 m/s
9.81 m/s²
0 m
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056112168224280018375573

Velocity Sweep

Watch how increasing launch speed scales the arc. With fixed axes you can directly compare how much smaller drag makes each trajectory.

50 m/s
45 °
9.81 m/s²
0 m
Loading chart...
056112168224280018375573

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