Interactive physics simulator
Range of a Projectile
Analyze how the initial speed, launch angle, elevation, and gravity govern the horizontal range. Try the target landing challenge, scrub the timeline, or compare complementary launch angles.
Horizontal Range Simulator
Configure variables to animate. Drag the chronometer scrubber to analyze horizontal displacement over time.
Live Result
- Initial Speed (v_0)
- 15 m/s
- Launch Angle (θ)
- 45°
- Platform Height (h)
- 0 m
- Horizontal Range (R)
- 22.9 m
- Time of Flight (T)
- 2.16 s
- Elapsed Time (t)
- 0.00 s
- Position (x, y)
- (0.0, 0.0) m
- Horizontal Speed (v_x)
- 10.6 m/s
- Vertical Speed (v_y)
- 0.0 m/s
- Gravity (g)
- 9.8 m/s²
- Active Equation
- R = v0²·sin2θ / g
What is the Range of a Projectile?
The Horizontal Range (denoted by R) is the total horizontal distance covered by a projectile from its launch point until it returns to its launch height (or hits the ground).
In an ideal physics model where air resistance is neglected, no forces act horizontally on the projectile. Therefore, the horizontal speed remains constant ($v_x = v_0 \cos\theta$) throughout the entire trajectory. The range is simply the product of this constant horizontal velocity and the total time of flight ($T$): **R = vx × T**.
Key Range Principles
Core concepts that determine how far a projectile will land.
- Maximum Range at 45°: When launching from ground level, a launch angle of exactly 45° achieves the maximum horizontal distance.
- Symmetric Complementary Angles: On level ground, launch angles that are complementary (add up to 90°, e.g., 30° and 60°) result in the exact same range, though their heights and flight times differ.
- Elevation Decreases Peak Angle: Firing from an elevated cliff changes the landing physics. The optimal angle for maximum range drops below 45°.
- Mass Independence: Without air resistance, heavy objects and light objects land at the same distance if launched with the same velocity and angle.
Range Formulas
R = v0x × T
v0x = v0 × cos(θ)
Depending on launch parameters, the range formula simplifies:
- Level Ground Launch (h = 0):
R = v02 × sin(2θ)g - Horizontal Launch off Cliff (θ = 0°):
R = v0 × √(2hg) - General elevated launcher with height h:
R = v0cos(θ) × v0sin(θ) + √((v0sin(θ))2 + 2gh)g
Symmetry and Launch Variations
| Launch Angle θ | Level Ground Range | Flight Time (T) | Physical Description |
|---|---|---|---|
| 45° (Optimal) | Maximum Range | Moderate | Balances flight time and horizontal speed perfectly. |
| 30° (Shallow) | Same as 60° | Shorter | Fired low; reaches target quickly but has small fall time. |
| 60° (High lob) | Same as 30° | Longer | Fired high; spends much time climbing but has low speed. |
| 0° (Horizontal) | R = v₀√(2h/g) | Shortest | Starts with zero upward speed. Height h determines time. |
| 90° (Vertical) | 0 meters | Longest | Fires straight up, landing directly back on the pad. |
Complementary Angle Symmetry
Observe how trajectories launched at complementary angles (30° and 60°) land at the identical point, while 45° lands farthest:
Solved Examples
A golfer hits a golf ball with an initial velocity of 40 m/s at an angle of 45° above the horizontal. Calculate the horizontal range of the ball. Use g = 9.8 m/s².
- Identify the given values: initial speed v0 = 40 m/s, launch angle θ = 45°, and gravity g = 9.8 m/s².
- Since the launch and landing are at the same level, use the level-ground range formula: R = v0² × sin(2θ) / g.
- Calculate the trigonometric component: sin(2 × 45°) = sin(90°) = 1.
- Substitute values into the equation: R = (40)² × 1 / 9.8 = 1600 / 9.8 ≈ 163.27 meters.
Answer: Horizontal Range ≈ 163.27 m
A stone is thrown horizontally with a speed of 10 m/s off a cliff that is 20 m high. Find the horizontal distance from the base of the cliff where it strikes the ground. Use g = 9.8 m/s².
- Identify the given values: launch height h = 20 m, initial speed v0 = 10 m/s with horizontal direction (θ = 0°), and gravity g = 9.8 m/s².
- For a horizontal launch, the range formula is: R = v0 × √(2h/g).
- Calculate the time of flight component: T = √(2 × 20 / 9.8) = √(40 / 9.8) ≈ 2.02 seconds.
- Substitute time and speed to find range: R = 10 m/s × 2.02 s ≈ 20.20 meters.
Answer: Horizontal Range ≈ 20.20 m
A launcher on an elevated platform 15 m high fires a projectile at 20 m/s at an angle of 30° above the horizontal. Calculate the final horizontal range. Use g = 9.8 m/s².
- Identify known values: launch height h = 15 m, initial speed v0 = 20 m/s, angle θ = 30°, and gravity g = 9.8 m/s².
- Decompose the initial velocity components: v0x = v0 × cos(30°) = 20 × 0.866 = 17.32 m/s; v0y = v0 × sin(30°) = 20 × 0.50 = 10.00 m/s.
- Find the total flight time T by solving the vertical quadratic equation: y = h + v0yt - ½gt² = 0, which gives: 15 + 10T - 4.9T² = 0, or 4.9T² - 10T - 15 = 0.
- Using the quadratic formula: T = (10 + √(10² - 4 × 4.9 × (-15))) / 9.8 = (10 + √(100 + 294)) / 9.8 = (10 + √394) / 9.8 ≈ 3.05 seconds.
- Calculate the horizontal range using flight time: R = v0x × T = 17.32 m/s × 3.05 s ≈ 52.83 meters.
Answer: Horizontal Range ≈ 52.83 m
Common Mistakes
- Using the level-ground range formula on cliffs: Substituting parameters into v0²sin(2θ)/g when launching from a height (h > 0). You must calculate total flight time including the platform drop height first.
- Thinking complementary angles share flight times: Believing a 60° launch and a 30° launch land at the same place because they spent the same time in the air. The 60° launch flies much higher and stays in the air longer, compensating for its slower horizontal speed.
- Mixing sin(2θ) with sin²(θ): Confusing the range angle function (sin(2θ)) with the maximum height angle function (sin²(θ)).
- Assuming mass affects range: Thinking heavier projectiles land closer. Mass does not appear anywhere in ideal kinematics math.
Quick Summary
- Horizontal range is the total horizontal distance a projectile travels.
- Level ground formula: R = v02 × sin(2θ) / g.
- Maximum range occurs at 45° for ground-level launches.
- Complementary launch angles (e.g. 30° and 60°) achieve the identical range on level ground.
- Height increases the horizontal range by prolonging the flight duration.
- Optimal angle drops below 45° when the launch platform height h > 0.
Practice Questions
1. A soccer ball is kicked from ground level at 22 m/s at an angle of 45°. What is its horizontal range? (Use g = 9.8 m/s²)
Using R = v0²sin(2θ) / g. Since θ = 45°, sin(2θ) = sin(90°) = 1. R = (22)² × 1 / 9.8 = 484 / 9.8 ≈ 49.39 meters.
2. An arrow is shot horizontally at 35 m/s from a castle wall 5 meters above the ground. How far does it travel horizontally? (Use g = 10 m/s²)
T = √(2h/g) = √(2 × 5 / 10) = √(10/10) = 1.00 second. R = v0 × T = 35 × 1.00 = 35.00 meters.
3. Compare the ranges of two baseballs launched at 30° and 60° on a flat field with the same initial velocity of 25 m/s. (Use g = 9.8 m/s²)
Complementary angles (θ₁ + θ₂ = 90°) achieve the exact same horizontal range on level ground. R = v0²sin(2θ) / g = 25² × sin(60°) / 9.8 ≈ 625 × 0.866 / 9.8 ≈ 55.23 meters for both.
4. A spring launcher fires a steel marble at 12 m/s on a table. What launch angle(s) will achieve a horizontal range of exactly 11 meters? (Use g = 9.8 m/s²)
R = v0²sin(2θ)/g ⇒ 11 = 12² × sin(2θ) / 9.8 ⇒ sin(2θ) = (11 × 9.8) / 144 = 107.8 / 144 ≈ 0.7486. Thus, 2θ ≈ arcsin(0.7486) ≈ 48.47° or 131.53°, yielding launch angles θ ≈ 24.2° or 65.8°.
5. If gravity is reduced by half, how does the horizontal range of a projectile fired with the same speed and angle change?
Since R is inversely proportional to gravity (R ∝ 1/g), cutting the gravitational acceleration in half will exactly double the horizontal range.
6. A projectile launched at 45° achieves a range of 80 m. What is its range if the launch speed is doubled? (Assume level ground)
Range is proportional to the square of launch speed (R ∝ v0²). Doubling the launch speed increases the range by a factor of 4 (2² = 4). The new range will be 80 × 4 = 320 meters.
FAQ
Frequently Asked Questions
What is the range of a projectile?
The horizontal range of a projectile is the total horizontal displacement from the launch position to the landing position, where the projectile hits the ground or platform height.
What is the formula for projectile range on level ground?
For a launch and landing at the exact same height, the formula is: R = v0² × sin(2θ) / g, where v0 is the initial speed, θ is the launch angle, and g is gravity.
Why does 45° yield the maximum range on level ground?
The range equation contains sin(2θ). The sine function reaches its maximum value of 1 when its argument is 90°. Setting 2θ = 90° gives θ = 45°. This launch angle perfectly balances vertical flight time and horizontal speed.
What are complementary launch angles?
Complementary launch angles are pairs of angles that add up to 90° (such as 30° and 60°, or 15° and 75°). When fired at the same speed on level ground, complementary angles result in the exact same horizontal range.
Do complementary angles stay in the air for the same amount of time?
No. Even though they land at the same horizontal distance, the higher angle (e.g. 60°) reaches a higher peak height and stays in the air longer than the shallower angle (e.g. 30°).
How does launching from an elevated cliff affect the optimal launch angle?
When launching from an elevated platform (h > 0), the optimal angle for maximum range is slightly less than 45° (typically between 35° and 42° depending on height). A lower angle provides more horizontal velocity component to cover distance during the extended drop time.
What is the range formula for a horizontal launch off a tower?
For a launch angle θ = 0°, the horizontal range is: R = v0 × √(2h/g), where h is the tower height and v0 is the horizontal launch speed.
Does the mass of the projectile affect the range?
No. In a vacuum model without air resistance, all objects accelerate downward at the same rate due to gravity regardless of their mass. Therefore, mass has no effect on the trajectory or range.
How does gravity affect the range?
Range is inversely proportional to gravity (R ∝ 1/g). Firing a projectile on the Moon (g = 1.62 m/s²) will result in a range about 6 times farther than on Earth (g = 9.8 m/s²) under identical launch conditions.
Can a projectile have a range of zero?
Yes. A projectile launched straight upward (90°) or straight downward (0° from platform edge with no speed) will have a horizontal range of zero, as it has no horizontal velocity component.
How does air resistance affect horizontal range in the real world?
Air resistance exerts a drag force opposing the velocity vectors. This continuously slows down the horizontal speed and shortens the flight time, causing the actual range to be significantly shorter than the vacuum math prediction.
What is the relationship between launch speed and range?
Range is proportional to the square of initial speed (R ∝ v0²). Doubling the launch velocity will quadruple (4x) the horizontal range of the projectile.