Projectile Motion Simulation
Example 4.9: Cricket Ball Projectile
Maximum Height:
10.0 m
Time of Flight:
2.9 s
Horizontal Range:
69 m
Current Height:
0.00 m
Current Distance:
0.00 m
Time Elapsed:
0.00 s
Velocity Vector
Acceleration Vector
Explanation:
This simulation demonstrates projectile motion based on Example 4.9 from Class 11 Physics. A cricket ball is thrown at 28 m/s at a 30° angle above the horizontal.
Key Concepts:
- Maximum Height: h = (v₀ sinθ)²/(2g) = (28×sin30°)²/(2×9.8) = 10.0 m
- Time of Flight: T = (2v₀ sinθ)/g = (2×28×sin30°)/9.8 = 2.9 s
- Horizontal Range: R = (v₀² sin2θ)/g = (28²×sin60°)/9.8 ≈ 69 m
- The horizontal motion has constant velocity (no acceleration)
- The vertical motion has constant acceleration due to gravity (9.8 m/s² downward)
