Projectile Motion
Why a thrown ball traces a perfect parabola — and how to predict exactly where it lands.
The glowing arc is illustrative — the real, physically-exact path is in the simulation below.
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The big idea: A projectile is anything thrown, launched, or dropped that then moves under gravity alone. The single most powerful idea in this whole lesson: you can split the motion into two independent problems — horizontal and vertical — and solve each on its own.
The idea that makes projectiles easy
Watch a basketball sail toward the hoop. It feels like one smooth, curved motion — and it is. But here's the insight that makes projectiles simple: the horizontal and vertical motions do not affect each other.
Horizontally, nothing pushes the ball (we ignore air resistance), so it keeps a constant speed. Vertically, gravity pulls it down, speeding its fall and slowing its rise — exactly as if you had simply dropped it. Run both at once and the curve you get is a parabola.
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The core principle
Horizontal and vertical motion are independent. A bullet fired horizontally and a bullet dropped from the same height hit the ground at the same time — gravity treats them identically.
Writing it as equations
Split the launch speed \(v_0\) at angle \(\theta\) into components, then apply constant-velocity motion horizontally and constant-acceleration motion (\(g \approx 9.8\,\text{m/s}^2\) downward) vertically:
\[ v_{0x} = v_0\cos\theta \qquad v_{0y} = v_0\sin\theta \]
Step 1 — break the launch velocity into components.
\[ x(t) = v_0\cos\theta \; t \]
Horizontal position: constant speed, so distance grows linearly with time.
\[ y(t) = v_0\sin\theta \; t - \tfrac{1}{2} g t^2 \]
Vertical position: rises, slows, stops, falls — the −½gt² term is gravity.
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Why a parabola, exactly?
Solve the horizontal equation for time, \(t = x /(v_0\cos\theta)\), and substitute into the vertical one. You get \(y\) as a function of \(x\) with an \(x^2\) term — the definition of a parabola. The shape isn't a coincidence; it falls straight out of the algebra.
🎮 Interactive: Launch a projectile LIVE
Drag the sliders for launch speed and angle. Watch the trajectory, the velocity components, and the live range/height numbers update. Try to find the angle that gives maximum range.
The results worth knowing (and re-deriving)
From those two position equations you can pull out three headline quantities. None of them is worth memorizing blindly — each falls out of \(x(t)\) and \(y(t)\) in a line or two of algebra. Play with the simulation above and confirm each one for yourself:
\[ t_{\text{flight}} = \frac{2 v_0 \sin\theta}{g} \]
Time of flight (for level ground).
\[ H = \frac{v_0^2 \sin^2\theta}{2g} \]
Maximum height reached at the peak.
\[ R = \frac{v_0^2 \sin 2\theta}{g} \]
Horizontal range. Because sin2θ peaks at 2θ = 90°, range is maximum at θ = 45°.
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When these three formulas apply
These tidy results assume level ground (launch height = landing height), no air resistance, and a constant \(g\). Launch off a cliff or into a stiff wind and the shortcuts break — but the method doesn't. Go back to the position equations \(x(t)\) and \(y(t)\) and you can still solve it.
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The 45° surprise (and a hidden symmetry)
Range depends on \(\sin 2\theta\), which is largest at \(\theta = 45°\). A neat consequence: complementary angles give the same range — a 30° and a 60° launch land in the same spot (one is a high lob, the other a flat drive). Verify it in the simulation!
📝 Worked example: A ball is kicked at 20 m/s at 30° above level ground. How far does it travel horizontally before landing? (Use g = 9.8 m/s².)
- Use the range formula: \(R = \dfrac{v_0^2 \sin 2\theta}{g}\).
- Here \(v_0 = 20\), \(\theta = 30°\) so \(2\theta = 60°\), and \(\sin 60° \approx 0.866\).
- \(R = \dfrac{20^2 \times 0.866}{9.8} = \dfrac{400 \times 0.866}{9.8} = \dfrac{346.4}{9.8}\).
✓ R ≈ 35.3 m.
📝 Worked example: Now solve that same 20 m/s, 30° kick from first principles — no memorized range formula. Find the time of flight, then the range.
- Resolve the launch velocity: \(v_{0x} = 20\cos30° = 17.3\,\text{m/s}\) and \(v_{0y} = 20\sin30° = 10\,\text{m/s}\).
- The ball lands when \(y = 0\): \(v_{0y}t - \tfrac{1}{2}gt^2 = 0 \Rightarrow t\,(10 - 4.9\,t) = 0\).
- Discard \(t = 0\) (the launch): \(t = 10 / 4.9 = 2.04\,\text{s}\) — the time of flight.
- Range is just horizontal speed × time: \(x = v_{0x}\,t = 17.3 \times 2.04\).
✓ R ≈ 35.3 m — identical to the shortcut formula, but derived from scratch. That's the general method that also works off a cliff.
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A common trap
At the very top of the arc the ball's vertical velocity is zero — but its horizontal velocity is not! The ball never stops moving; it just momentarily stops climbing. It's tempting to say "velocity = 0 at the top" — but only the vertical component is zero. The ball is still sailing forward the whole time.
Where you meet this in the real world
- Sport: the arc of a basketball, a golf drive, a long jump.
- Engineering: water from a fountain or fire hose, ballistics, the safe landing zone under a ski jump.
- Space: an orbit is really just a projectile falling so fast it keeps missing the Earth — the bridge to the Astrophysics module.
Check your understanding
1. A ball is launched at 15 m/s at 40° above level ground. Using R = v₀²sin2θ/g with g = 9.8 m/s², its range is closest to…
R = 15² × sin80° / 9.8 = 225 × 0.985 / 9.8 ≈ 22.6 m. (sin80° ≈ 0.985.)
2. You double the launch speed but keep the same angle and level ground. The range becomes…
Range depends on v₀² (R = v₀²sin2θ/g). Doubling v₀ multiplies the range by 2² = 4 — this is why a small speed boost matters so much.
3. At the highest point of a projectile's arc, its velocity is…
The vertical component is momentarily zero at the peak, but the horizontal component keeps its constant value for the whole flight.
4. On level ground, a 25° launch lands a certain distance away. Which other angle (same speed) lands at the exact same spot?
Complementary angles θ and (90° − θ) share the same range, because sin2θ = sin(180° − 2θ). Here 90° − 25° = 65° — a high lob and a flat drive that land together.
✅ Key takeaways
- Horizontal and vertical motion are independent — solve each axis on its own.
- Position: x = v₀cosθ·t and y = v₀sinθ·t − ½gt². Combine them and y-vs-x is a parabola.
- On level ground: range R = v₀²sin2θ/g, max height H = v₀²sin²θ/2g, flight time t = 2v₀sinθ/g.
- Range is greatest at 45°; complementary angles (like 30° & 60°) share the same range.