Projectile Motion Explained — Trajectories, Range and Height

Why thrown objects follow a curved path, the independence of horizontal and vertical motion, range and height formulas, and worked examples.

Quick answer

Projectile motion is the curved path of an object launched into the air that then moves under gravity alone. Its horizontal velocity stays constant while its vertical velocity changes at g = 9.81 m/s².

R=v02sin⁡2θg,H=(v0sin⁡θ)22gR = \frac{v_0^2\sin 2\theta}{g},\qquad H = \frac{(v_0\sin\theta)^2}{2g}

Unit: Range and height in meters (m), angles in degrees

Example: A ball launched at 20 m/s at 30° travels R = 400 × sin60° / 9.81 ≈ 35.3 m on flat ground.

What you will learn

  • Explain why a projectile's horizontal and vertical motions are independent
  • Split a launch velocity into horizontal and vertical components
  • Use the range, height and flight-time formulas
  • Predict how launch angle changes range and maximum height

What counts as a projectile

A projectile is any object that is launched and then left to move under gravity alone — a thrown ball after it leaves the hand, a cannonball after it leaves the barrel, water from a fountain. The key idea is that the launch force is over; from that moment, the only force acting is weight pulling straight down.

Air resistance is ignored in the standard treatment, which is a good approximation for heavy, slow objects over short flights and a poor one for feathers, shuttlecocks or high-speed artillery.

Horizontal and vertical motion are independent

This is the single most important idea in the topic. Gravity acts only vertically, so the horizontal component of velocity never changes (no horizontal force), while the vertical component changes at $g = 9.81\ \mathrm{m/s^2}$ downward — exactly like a ball dropped from rest.

Famously, a bullet fired horizontally and a bullet dropped from the same height hit the ground at the same time. The horizontal motion adds no help or hindrance to the fall; the two motions simply coexist.

Split the launch velocity v0v_0 at angle θ\theta above the horizontal using trigonometry:

v0x=v0cos⁡θ,v0y=v0sin⁡θv_{0x} = v_0 \cos\theta,\qquad v_{0y} = v_0 \sin\theta

Diagram of a projectile's curved parabolic path showing constant horizontal velocity arrows and changing vertical velocity arrowsv₀θh maxR (range)
The parabolic trajectory of a projectile: horizontal motion is uniform, vertical motion is accelerated.

Time of flight and maximum height

The vertical motion is just constant-acceleration motion with a=−ga = -g. Taking upward as positive and launching from the ground, the vertical velocity is vy=v0sin⁡θ−gtv_y = v_0\sin\theta - gt.

The projectile reaches its peak when vy=0v_y = 0, giving a time to the top of tup=v0sin⁡θ/gt_{up} = v_0\sin\theta / g. On flat ground the flight is symmetric, so the total flight time is twice that:

T=2v0sin⁡θgT = \frac{2v_0\sin\theta}{g}

The maximum height comes from vy2=v0y2−2gHv_y^2 = v_{0y}^2 - 2gH with vy=0v_y = 0 at the top:

H=(v0sin⁡θ)22gH = \frac{(v_0\sin\theta)^2}{2g}

The range formula

Range is the horizontal distance traveled. Since horizontal velocity is constant, range is simply horizontal speed times flight time:

R=v0xT=v0cos⁡θ⋅2v0sin⁡θg=v02sin⁡2θgR = v_{0x}T = v_0\cos\theta \cdot \frac{2v_0\sin\theta}{g} = \frac{v_0^2\sin 2\theta}{g}

where the identity 2sin⁡θcos⁡θ=sin⁡2θ2\sin\theta\cos\theta = \sin 2\theta was used. This formula assumes launch and landing at the same height and no air resistance.

Because sin⁡2θ\sin 2\theta peaks at 11 when 2θ=90∘2\theta = 90^\circ, maximum range occurs at a launch angle of $45^\circ$. Angles symmetric about 45∘45^\circ — like 30∘30^\circ and 60∘60^\circ — give identical ranges.

Why 45 degrees

At 45°, the launch velocity is split evenly between horizontal speed (which carries the projectile far) and vertical speed (which keeps it airborne). Tilting higher trades range for height; tilting lower trades airtime for speed across the ground.

Worked example: the full trajectory

A soccer ball is kicked at v0=20 m/sv_0 = 20\ \mathrm{m/s} at 30∘30^\circ above the horizontal. Find the range, maximum height and flight time.

First split the velocity: v0x=20cos⁡30∘=17.32 m/sv_{0x} = 20\cos 30^\circ = 17.32\ \mathrm{m/s}, v0y=20sin⁡30∘=10 m/sv_{0y} = 20\sin 30^\circ = 10\ \mathrm{m/s}.

Flight time: T=2(10)/9.81=2.04 sT = 2(10)/9.81 = 2.04\ \mathrm{s}.

Maximum height: H=102/(2×9.81)=5.10 mH = 10^2/(2 \times 9.81) = 5.10\ \mathrm{m}.

Range: R=v0xT=17.32×2.04=35.3 mR = v_{0x}T = 17.32 \times 2.04 = 35.3\ \mathrm{m} — matching the direct formula 400sin⁡60∘/9.81=35.3 m400\sin 60^\circ/9.81 = 35.3\ \mathrm{m}.

Launching from a height

The neat formulas above assume landing at launch height. When a projectile is launched from a cliff or a raised platform, the flight lasts longer because the object has extra distance to fall — so the range is larger than v02sin⁡2θ/gv_0^2\sin 2\theta/g predicts.

The fix is to solve the vertical motion separately: with launch height h0h_0, the landing time solves h0+v0yt−12gt2=0h_0 + v_{0y}t - \tfrac{1}{2}gt^2 = 0 (take the positive root), then multiply by the constant horizontal speed v0xv_{0x}. There is no single compact range formula for this case, which is why textbooks emphasize the flat-ground version.

Horizontal launch

An object launched horizontally (θ = 0) has v₀y = 0, so it falls exactly like a dropped object — flight time t = √(2h₀/g), range R = v₀t. The 'dropped vs fired bullet' demonstration is this special case.

Where the ideal model breaks down

Air drag reshapes real trajectories: it shortens range, lowers the peak, and makes the descent steeper than the ascent. For a golf ball, drag and spin (the Magnus effect) matter so much that the vacuum optimum of 45∘45^\circ drops to roughly 30∘30^\circ.

At very high speeds or long ranges, two more corrections appear: the curvature of the Earth changes what 'flat ground' means, and the Coriolis effect deflects long-range artillery. For classroom-scale problems, though, the no-drag parabola is accurate and is the model examiners expect.

Variables and units

SymbolNameUnitMeaning
v0v₀Launch speedm/sSpeed at the moment of launch
θθLaunch angledegreesAngle of launch above the horizontal
RRRangemHorizontal distance from launch to landing
HHMaximum heightmPeak vertical height above the launch level
TTTime of flightsTotal time the projectile is airborne
ggGravitational accelerationm/s²9.81 m/s² downward near Earth's surface

Common mistakes

✗ A faster horizontal launch makes the object hit the ground sooner.

Horizontal and vertical motions are independent. At the same launch height, a fast and a slow horizontal launch land at the same instant.

✗ Maximum range always occurs at 45°.

Only in a vacuum with launch and landing at equal heights. Air drag lowers the real optimum, and launching from a height favors a flatter angle.

✗ The velocity at the top of the trajectory is zero.

Only the vertical component is zero at the peak. The horizontal component is unchanged throughout the flight, so the projectile is still moving forward.

Real-world applications

  • Firefighting monitors are aimed using projectile equations to place water streams on upper floors.

  • Sports science uses launch-angle analysis to optimize javelin, shot put and long-jump technique.

  • Irrigation engineers space sprinkler heads using the range of water jets at given pressures.

  • Video-game physics engines integrate the same equations to simulate thrown objects and ballistics.

Frequently asked questions

Why is the trajectory a parabola?

Horizontal position grows linearly with time (x = v₀ₓt) while vertical position is quadratic (y = v₀ᵧt − ½gt²). Eliminating t between them gives y as a quadratic function of x — the definition of a parabola. This holds exactly when air resistance is ignored.

Do 30° and 60° really give the same range?

Yes, in the ideal model. Since sin(2×30°) = sin60° and sin(2×60°) = sin120° = sin60°, the range formula gives identical results. The higher angle just takes longer and reaches greater height — the classic high arc vs flat drive.

What happens if I launch straight up at 90°?

The range formula gives sin180° = 0, so the range is zero — the projectile goes straight up and lands back at the launch point (ignoring wind). Maximum height becomes v₀²/2g, the largest height any launch angle can give.

How does air resistance change the answers?

Drag opposes the velocity, so it reduces range, height and flight time, and the path is no longer a symmetric parabola — the descent is steeper than the ascent. Light, fast objects like shuttlecocks are affected far more than dense, slow ones like shot puts.