PHY005 · Mechanics · Classes 11–12
Projectile motion
Objective: To study projectile motion and find the angle that gives the maximum range.
🧰 Apparatus
- Spring launcher with angle scale
- Small ball
- Measuring tape
- Carbon paper
📐 Formulas and SI units
R = u² sin 2θ / g (m)
H = u² sin²θ / 2g (m)
T = 2u sin θ / g (s)
💡 Theory
Without air resistance, the horizontal velocity stays constant and the vertical motion has constant acceleration g. The path is a parabola. Range R = u² sin 2θ / g is largest at 45°, and complementary angles (30° and 60°) give the same range.
🪜 Procedure
- Fix the launcher at 15° and fire the ball onto carbon paper on the floor.
- Measure the range from the launch point.
- Repeat at 30°, 45°, 60° and 75°, three shots each.
- Plot range against angle.
- Compare 30° with 60° and find the maximum.
📋 Observations Simulated values
✅ Result and conclusion
At 20 m/s and 45°, the range is 40.8 m and the height 10.2 m; 45° gives the maximum range.
🔬 In a real lab
Air resistance shortens real ranges and moves the best angle below 45°; the launch speed varies a little between shots.
⚠️ Safety
Launch only soft balls, away from people and windows; never aim at anyone.
❓ Quick check
1. Which angle gives the maximum range on level ground?
2. At the top of the path the velocity is…
3. Which pairs of angles give equal ranges?
🏋️ Practice
Easyu = 10 m/s, θ = 45°. R?
Answer: ≈ 10.2 m
Mediumu = 20 m/s, θ = 30°. Time of flight?
Answer: ≈ 2.04 s
ChallengeMax height for u = 14 m/s straight up?
Answer: 10 m