A basketball arcs toward the hoop. A cannonball launches from a cliff. A long-jumper leaves the ground at an angle. All are examples of projectile motion — motion under gravity alone, with no other forces acting. The key insight is that horizontal and vertical motion are completely independent: the horizontal component travels at constant speed, while the vertical component accelerates at g = 9.8 m/s² downward. Treating them separately turns every projectile problem into two straightforward SUVAT calculations sharing one variable: time.
Horizontal: x = u cos θ × t (constant velocity, a = 0)
Vertical: y = u sin θ × t − ½gt² (SUVAT with a = −g)
Range: R = u² sin 2θ / g
Max height: H = (u sin θ)² / (2g)
Time of flight: T = 2u sin θ / g
u = launch speed (m/s) | θ = launch angle | g = 9.8 m/s²
The Two-Component Approach
Resolve the initial velocity u at angle θ into components:
Horizontal motion (no air resistance → no horizontal force → no horizontal acceleration):
Vertical motion (gravity acts downward; taking upward as positive, a = −g):
The link between components is time t — horizontal and vertical motion share the same time variable. This is the coupling that makes projectile problems solvable: find t from the vertical equations, then use it in the horizontal equation to find range.
Deriving the Range Formula
For a projectile launched from and landing at the same height (y = 0 on return):
Substitute into x = u cos θ × T:
Maximum range occurs at θ = 45°, where sin 2θ = sin 90° = 1 gives R_max = u²/g.
Deriving Maximum Height
At maximum height, vertical velocity = 0:
Time to reach max height = T/2 = u sin θ / g (by symmetry — same time up as down).
Worked Example 1: Classic Range Problem
A ball is kicked at 20 m/s at 30° above horizontal. Find: (a) time of flight; (b) maximum height; (c) horizontal range.
Components: u_x = 20 cos 30° = 17.32 m/s; u_y = 20 sin 30° = 10.0 m/s
(a) Time of flight:
(b) Maximum height:
(c) Range:
Or using the formula: R = u² sin 2θ / g = 400 × sin 60° / 9.8 = 400 × 0.866 / 9.8 = 35.3 m ✓
Worked Example 2: Projectile Launched from a Height
A ball is thrown horizontally at 15 m/s from a cliff 45 m high. Find: (a) time to hit the ground; (b) horizontal distance from base of cliff; (c) speed at impact.
Horizontal: u_x = 15 m/s, u_y = 0 (thrown horizontally)
(a) Vertical: y = ½gt² → 45 = ½ × 9.8 × t² → t² = 9.18 → t = 3.03 s
(b) x = u_x × t = 15 × 3.03 = 45.5 m
(c) v_x = 15 m/s (unchanged); v_y = gt = 9.8 × 3.03 = 29.7 m/s
Worked Example 3: Finding Launch Angle for a Target
A ball must travel 80 m horizontally and land at the same height, launched at 30 m/s. Find the launch angle(s).
Two angles work: a low, fast trajectory (30.3°) and a high, slow one (59.7°). Both give the same range because sin 2(30.3°) = sin 2(59.7°) — complementary angles give equal ranges.
Worked Example 4: Projectile with Non-Zero Landing Height
A ball is launched at 25 m/s at 40° above horizontal from ground level. A wall 20 m away is 5 m high. Does the ball clear the wall?
Time to reach x = 20 m: t = x/u_x = 20/(25 cos 40°) = 20/19.15 = 1.045 s
Height at that time: y = u_y × t − ½gt² = 25 sin 40° × 1.045 − ½ × 9.8 × 1.045²
The ball is 11.44 m high at the wall — it clears the 5 m wall with 6.44 m to spare.
Why Projectile Motion is a Parabola
The trajectory y(x) is a parabola. From x = u_x t → t = x/u_x. Substituting into y = u_y t − ½gt²:
This is y = ax − bx² — the standard form of a parabola. The coefficient of x gives the initial slope (tan θ); the coefficient of x² gives the curvature (determined by g and horizontal speed). This is explored further in the why is projectile motion a parabola article.
Effect of Launch Angle on Range and Height
| Angle θ | sin 2θ | Range (u=20 m/s) | Max height (u=20 m/s) |
|---|---|---|---|
| 15° | 0.500 | 20.4 m | 1.35 m |
| 30° | 0.866 | 35.3 m | 5.10 m |
| 45° | 1.000 | 40.8 m (MAX) | 10.2 m |
| 60° | 0.866 | 35.3 m | 15.3 m |
| 75° | 0.500 | 20.4 m | 19.1 m |
Complementary angles (30° and 60°, 15° and 75°) give equal ranges but different heights and flight times. The 45° launch maximises range; steeper angles maximise height at the cost of range.
Real-World Complications
Air resistance: Real projectiles experience drag proportional to v² (at high speeds). This shortens range, reduces max height, and breaks the symmetry of the trajectory — the descending path is steeper than the ascending path. The optimal angle for maximum range with air resistance is less than 45° (typically 30–40° for a sports ball). Long-range artillery accounts for drag, wind, Earth's rotation (Coriolis effect), and even the curvature of the Earth.
Spin: A spinning ball experiences the Magnus effect — a force perpendicular to both the spin axis and velocity. This is why a spinning football curves mid-flight (a "banana" free kick), why a topspin tennis shot dips faster, and why a backspin golf shot stays in the air longer.
Sports applications: The ideal basketball free throw (distance ~4.57 m, basket height 3.05 m, release height ~2.13 m) requires a minimum launch speed when thrown at about 51° and benefits from backspin which increases the effective target size. Olympic shot put records require careful optimisation of launch angle (~42°) accounting for the non-zero release height (~2.1 m above ground).
Using the Projectile Motion Calculator
The projectile motion calculator solves all these equations instantly — enter launch speed and angle, get range, max height, time of flight, and the full trajectory plotted automatically. It also works for any planet's gravity, so you can compare a football kick on Earth versus the Moon.
Common Mistakes
Using the full launch speed instead of components. The horizontal and vertical equations each need their respective components (u cos θ and u sin θ), not the total speed u. Plugging u directly into vertical equations is the most common projectile error.
Forgetting to take upward as positive consistently. If upward is positive, g = −9.8 m/s² in all vertical equations. If downward is positive, g = +9.8 m/s² and initial upward velocities become negative. Pick one and stick to it for the entire problem.
Assuming 45° always gives maximum range. This is only true when launching and landing at the same height. When launched from above (e.g. off a cliff), the optimal angle is less than 45°. When the target is higher than the launch point, the optimal angle is greater than 45°.
Frequently Asked Questions
What is projectile motion?
Projectile motion is the motion of an object launched into the air that moves under gravity alone (no engine, no air resistance). The horizontal and vertical components of motion are independent: horizontally, velocity is constant (no force); vertically, the object accelerates at g = 9.8 m/s² downward. The path is a parabola. Any object thrown, kicked, shot, or dropped at an angle — a ball, bullet, long-jumper, water stream — undergoes projectile motion if air resistance is negligible.
What is the formula for the range of a projectile?
The horizontal range of a projectile launched at speed u and angle θ from and returning to the same height is R = u² sin 2θ / g. This gives maximum range at θ = 45° (where sin 2θ = sin 90° = 1), giving R_max = u²/g. Complementary angles (e.g. 30° and 60°) give equal ranges. If the launch and landing heights differ, use the component method: find time of flight from the vertical equation, then multiply by horizontal speed.
Why does a projectile launched at 45° have maximum range?
Range R = u² sin 2θ / g is maximised when sin 2θ = 1, i.e. 2θ = 90°, i.e. θ = 45°. At 45°, horizontal and vertical components are equal (u/√2 each), giving the optimal balance between time of flight (determined by vertical component) and horizontal speed. Lower angles have high horizontal speed but short flight time; higher angles have long flight time but low horizontal speed. 45° maximises their product — but only when launch and landing heights are equal.
Are horizontal and vertical motion truly independent in a projectile?
Yes — in the idealised case with no air resistance. Gravity acts only vertically, so it only affects vertical velocity. There is no horizontal force, so horizontal velocity stays constant throughout the flight. This independence is the key insight of projectile motion analysis. The two components share only the time variable. This was first demonstrated by Galileo, who showed that a ball dropped vertically and one launched horizontally from the same height hit the ground simultaneously — proof that horizontal motion doesn't affect the vertical.
How do you solve a projectile motion problem step by step?
Step 1: Resolve initial velocity into components — u_x = u cos θ (horizontal), u_y = u sin θ (vertical). Step 2: Set up the horizontal equation x = u_x t and vertical SUVAT equations with a = −g (taking upward positive). Step 3: Identify the unknown and which equation to use — usually find t from the vertical equation first, then use it in the horizontal. Step 4: Substitute and calculate. Step 5: Check the sign and magnitude make physical sense. For problems involving landing at a different height, set y = Δh in the vertical equation and solve the resulting quadratic for t.
What is the velocity of a projectile at maximum height?
At maximum height, the vertical component of velocity is exactly zero (v_y = 0) — this is the definition of the highest point (the object momentarily stops moving vertically before starting to fall). The horizontal component is unchanged throughout: v_x = u cos θ. So at maximum height, the speed equals u cos θ, directed horizontally. The time to reach maximum height is t = u sin θ / g, and the height is H = (u sin θ)² / (2g).
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