Theory — Projectile Motion
Two Motions at Once
A projectile is any object launched into the air and moving under gravity alone (we ignore air resistance). The key idea is that its motion splits into two independent parts: a horizontal motion at constant velocity, and a vertical motion with constant downward acceleration g. The two happen at the same time but do not affect each other — gravity changes only the vertical motion.
Vertical: vy = v·sin(θ) (changes by −g each second)
v = launch speed, θ = launch angle above horizontal, g = 9.81 m/s²
Time of Flight
The projectile rises, slows, stops vertically at the top, then falls. On level ground it lands when the vertical displacement returns to zero. The total time in the air is the time of flight.
Maximum Height and Range
The maximum height depends only on the vertical part of the launch. The range — the horizontal distance travelled before landing — is the constant horizontal velocity multiplied by the time of flight.
Range: R = v²·sin(2θ) / g
Complementary Angles Give the Same Range
Because the range depends on sin(2θ), and sin(2θ) has the same value for an angle and its complement (for example sin(60°) = sin(120°)), two launch angles that add up to 90° produce exactly the same range. So 30° and 60° land at the same horizontal distance — even though the 60° shot flies much higher and stays in the air longer. The maximum possible range occurs at 45°, halfway between any such pair.
Increase the angle
Higher launch angle → greater height and longer flight time, but range peaks at 45° then falls.
Increase the speed
Both range and height grow with the square of the launch speed (R, H ∝ v²).
Complementary pair
θ and (90° − θ) give equal range. 30° & 60° land together; 45° goes farthest.
| Quantity | Equation | Depends on |
|---|---|---|
| Horizontal velocity | v·cos(θ) | constant throughout flight |
| Initial vertical velocity | v·sin(θ) | decreases by g each second |
| Time of flight | 2v·sin(θ)/g | vertical motion only |
| Maximum height | v²·sin²(θ)/(2g) | vertical motion only |
| Range | v²·sin(2θ)/g | maximum at 45° |
Instructions — Running the Virtual Experiment
The Launch tab lets you fire the cannon at any angle and speed and read the range, height, and time directly. The Range vs Angle tab records range against launch angle so you can find the maximum and confirm the complementary-angle rule. Record every reading in your lab notebook.
Simulation — Launch the Projectile
Controls
Fixed speed: 20 m/s
Set angle
| Launch angle (°) | Range (m) |
|---|---|
| No readings yet — pick an angle and click "Record reading". | |
Team Questions
Example Lab Report
Sample report demonstrating the expected format and level of detail. Use as a guide for your own submission.
Determining the Height and Range of a Projectile
Physics | Section: [Your Section] | Date: [Date]
Lab Members: [Names of all members present]
Purpose
To investigate the motion of a projectile launched at various angles with a fixed initial speed, to measure the range and maximum height for each launch, and to compare the measured values with theoretical predictions. A particular aim is to examine how the range depends on the launch angle, including the special cases of the maximum-range angle and complementary angles.
Theory
A projectile undergoes constant-velocity horizontal motion and constant-acceleration vertical motion simultaneously and independently. For a launch on level ground with speed v at angle θ, the time of flight, maximum height, and range are given below. The range is maximised at 45°, and complementary angles produce equal ranges because the range depends on sin(2θ).
T = 2v·sin(θ)/g
H = v²·sin²(θ)/(2g)
R = v²·sin(2θ)/g (g = 9.81 m/s²)
Calculations — Sample: v = 20 m/s at 45°
Range: R = (20²·sin(90°))/9.81 = (400·1)/9.81 = 40.8 m
Maximum height: H = (20²·sin²(45°))/(2·9.81) = (400·0.5)/19.62 = 10.2 m
Time of flight: T = (2·20·sin(45°))/9.81 = (40·0.707)/9.81 = 2.88 s
Results Table — Range and Height (v = 20 m/s, g = 9.81 m/s²)
| Scenario | Angle | Range R (m) | Max height H (m) | Time T (s) |
|---|---|---|---|---|
| 1 | 45° | 40.8 | 10.2 | 2.88 |
| 2 | 60° | 35.3 | 15.3 | 3.53 |
| 3 | 30° | 35.3 | 5.1 | 2.04 |
Experimental values read from the simulation matched these theoretical values within a fraction of a metre.
Discussion
The measured ranges and heights agreed closely with the theoretical predictions, confirming the projectile equations. The 45° launch produced the greatest range (40.8 m), as expected since sin(2θ) reaches its maximum value of 1 at θ = 45°.
The most instructive comparison is between Scenario 2 (60°) and Scenario 3 (30°): both produced the same range of 35.3 m, even though the launches look very different. This happens because the range depends on sin(2θ), and sin(120°) = sin(60°), so complementary angles (angles that add to 90°) always give equal ranges. The two shots differ in their other properties: the 60° launch climbed much higher (15.3 m versus 5.1 m) and stayed in the air longer (3.53 s versus 2.04 s), because a steeper launch puts more of the speed into the vertical direction. The 30° launch had a larger horizontal velocity but less time aloft, and the two effects combine to give the same horizontal distance.
Conclusion
The experiment confirmed the equations for projectile range and maximum height and demonstrated that range is greatest at 45°. It also showed that complementary launch angles (30° and 60°) produce equal ranges, with the higher angle reaching a greater height and longer flight time. Theoretical and experimental values were in close agreement.
Practice Questions
Show all work and include units in your answers. Use g = 9.81 m/s² and ignore air resistance.