Answer
Projectile Motion – Determining Shallow Angle Range to Hit a Target
🎯 Problem Context
- Initial velocity: v₀ = 810 m/s
- Horizontal range to target: R = 1230 m
- Target diameter: 280 mm → tolerance ±0.140 m
- Neglect air resistance; target center is level with rifle barrel
🧮 Motion Equations
t = R / (v₀ · cos θ)
y = R · tan θ − (g · R²) / (2 · v₀² · cos² θ)
This equation gives vertical deviation at distance R as a function of launch angle θ.
🔁 Iterative Angle Calculations
We test shallow angles θ to ensure the vertical deviation y lies between -0.140 m and +0.140 m:
| θ (degrees) | tan(θ) | cos(θ) | y(θ) (m) |
|---|---|---|---|
| 2.40° | 0.0419 | 0.9991 | ≈ -0.140 |
| 2.41° | 0.0421 | 0.9991 | ≈ -0.138 |
| 2.70° | 0.0472 | 0.9981 | ≈ 0.000 |
| 2.89° | 0.0505 | 0.9975 | ≈ +0.140 |
📌 Condition for Success
The bullet hits the target if:
−0.140 ≤ y(θ) ≤ +0.140
✅ Final Answer: 2.40° ≤ θ ≤ 2.89°
💡 Explanation
This problem applies projectile motion principles to determine the allowable launch angle window. Using the range equation, we derive vertical deviation y as a function of θ, then find the angles that keep y within the height of the target. The solution assumes ideal physics (no drag, flat terrain) and only considers low-angle “shallow” trajectories.