A projectile is any object launched into the air and left to move under gravity alone — no engine, no rocket, no continued pushing. From a thrown ball to an artillery shell to a basketball shot, every such object follows the same elegant mathematics. Master three equations and you can predict exactly where anything will land, how high it will go, and how long it will take.

45°Optimal launch angle
3Key equations
g = 9.81m/s² downward
Real applications

1. The Physical Idea

The key insight in projectile motion — discovered by Galileo Galilei in the early 1600s — is that horizontal and vertical motion are completely independent of each other. Gravity acts only vertically. There is nothing acting horizontally (ignoring air resistance). So we treat the two directions separately and combine the results at the end.

  • Horizontally: No force acts → constant velocity → x = u·cos(θ)·t
  • Vertically: Gravity acts at constant g downward → uniform acceleration → y = u·sin(θ)·t − ½gt²
⚡ Galileo’s Principle

The horizontal and vertical components of projectile motion are completely independent. Gravity does not affect horizontal motion; horizontal velocity does not affect vertical fall.

This is why two balls — one dropped vertically, one launched horizontally from the same height — hit the ground at exactly the same time.


2. Decomposing Initial Velocity

When a projectile is launched at speed u and angle θ above the horizontal, the initial velocity breaks into two components:

u_x = u·cos(θ)     u_y = u·sin(θ)
Horizontal and vertical components of initial velocity

The horizontal component u_x remains constant throughout the flight. The vertical component u_y decreases at rate g (9.81 m/s²) due to gravity — reaching zero at maximum height, then becoming negative as the object falls back down.

Launch Anglecos(θ) — Horizontalsin(θ) — VerticalEffect
1.00 (maximum)0.00No vertical launch — hits ground instantly
30°0.8660.500Good range, moderate height
45°0.7070.707Maximum range (equal horizontal & vertical)
60°0.5000.866Same range as 30°, greater height
90°0.001.00 (maximum)Straight up — zero range, maximum height

3. The Three Key Equations

Maximum Height (H)

At maximum height, the vertical velocity is zero. Using v² = u_y² − 2gH:

H = u²sin²(θ) / 2g
Maximum height above launch point (assuming level ground)

Time of Flight (T)

Setting vertical displacement back to zero (for landing at the same height as launch):

T = 2u·sin(θ) / g
Total time from launch to landing

Horizontal Range (R)

Range = horizontal speed × time of flight. Using 2sin(θ)cos(θ) = sin(2θ):

R = u²·sin(2θ) / g
Total horizontal distance from launch to landing

4. Why 45° Gives Maximum Range

The range formula is R = u²·sin(2θ)/g. The only angle-dependent factor is sin(2θ). Since sin is maximum when its argument is 90°:

sin(2θ) is maximum when 2θ = 90°, i.e. θ = 45°.

At 45°: sin(90°) = 1, giving R_max = u²/g. Any other angle gives sin(2θ) < 1, meaning a shorter range.

🎯

Beautiful Symmetry: Angles that add to 90° give identical ranges. A 30° launch gives the same range as a 60° launch (since sin(60°) = sin(120°)). The low-angle shot arrives faster on a flatter trajectory; the high-angle shot arrives slower from above. Same landing spot, completely different paths.


5. Worked Examples

Example 1Complete projectile analysis

Problem: A ball is launched at 20 m/s at 30° above horizontal. Find maximum height, time of flight, and range. (g = 9.81 m/s²)

1
Maximum height: H = (20² × sin²30°) / (2 × 9.81) = (400 × 0.25) / 19.62 = 100 / 19.62 ≈ 5.10 m
2
Time of flight: T = 2 × 20 × sin30° / 9.81 = (2 × 20 × 0.5) / 9.81 = 20 / 9.81 ≈ 2.04 s
3
Range: R = 20² × sin60° / 9.81 = 400 × 0.866 / 9.81 = 346.4 / 9.81 ≈ 35.3 m
✓ H = 5.10 m, T = 2.04 s, R = 35.3 m. Note: sin(2 × 30°) = sin(60°) ≈ 0.866 used in range formula.
Example 2Finding launch speed from range

Problem: A golf ball lands 200 m away after being launched at 45°. What was the launch speed? (g = 9.81 m/s²)

1
R = u²·sin(2θ)/g → u² = Rg / sin(2θ)
2
u² = 200 × 9.81 / sin(90°) = 1962 / 1 = 1962 m²/s²
3
u = √1962 ≈ 44.3 m/s
✓ Launch speed ≈ 44.3 m/s (about 159 km/h) — a realistic long drive speed in golf.

6. Common Exam Mistakes

✗ Mistake 1

Using the full initial speed in vertical equations. Only u·sin(θ) drives vertical motion. Never substitute the full launch speed u into vertical kinematic equations — decompose first.

✗ Mistake 2

Applying the range formula when launch and landing heights differ. R = u²sin(2θ)/g is only valid when the projectile lands at the same height it was launched from. If it lands on a cliff, a slope, or below the launch point, you must use the component equations directly.

✗ Mistake 3

Assuming maximum height occurs at half the total time always. This is only true when the launch and landing heights are equal. If the projectile lands at a different height, maximum height occurs at T_up = u·sin(θ)/g, which is not necessarily T/2.

✗ Mistake 4

Forgetting that at maximum height, vertical velocity = 0, not total velocity = 0. At the peak, v_y = 0 but v_x = u·cos(θ) is still non-zero. The projectile is still moving horizontally — it is not momentarily stopped.


7. Frequently Asked Questions

Why does 45° give maximum range? +
Because the range formula R = u²sin(2θ)/g is maximised when sin(2θ) = 1, which occurs when 2θ = 90°, so θ = 45°. At 45° the launch balances horizontal speed and vertical height perfectly. Any other angle gives more of one and less of the other, reducing the total range.
Does air resistance affect projectile motion? +
Yes — significantly for real objects. Air resistance reduces range, maximum height, and time of flight. It also means the optimal angle for maximum range is less than 45° (typically around 30–40° for a golf ball, depending on speed). The equations in this article assume no air resistance (ideal projectile), which is an excellent approximation for dense, slow-moving objects over short distances.
What is the trajectory shape of a projectile? +
A perfect parabola (in the absence of air resistance). This follows mathematically from the independence of horizontal and vertical motion: x varies linearly with time; y varies quadratically with time. Eliminating t gives y as a quadratic function of x — which is the definition of a parabola. See our dedicated article Why Is Projectile Motion a Parabola? for the full proof.

Conclusion

Projectile motion is one of the most beautiful applications of Galileo’s independence principle. Horizontal and vertical motion are completely separate; treat them that way and the mathematics flows naturally. The three key equations — H = u²sin²θ/2g, T = 2usinθ/g, R = u²sin(2θ)/g — come directly from kinematics applied to each direction independently. Master these and you can analyse every thrown ball, every launched rocket, every kicked football with complete precision.