A projectile is launched with speed 20 m/s at 60 degrees. What is its horizontal range ignoring air resistance?

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Multiple Choice

A projectile is launched with speed 20 m/s at 60 degrees. What is its horizontal range ignoring air resistance?

Explanation:
The key idea is the projectile range without air resistance, given by R = v^2 sin(2θ) / g. With speed 20 m/s and launch angle 60°, compute sin(2θ) = sin(120°) = sin(60°) = √3/2 ≈ 0.866. Then R = (20^2) × 0.866 / 9.8 = 400 × 0.866 / 9.8 ≈ 346.4 / 9.8 ≈ 35.4 m. So the horizontal range is about 35 m, which matches the option described as approximately 35 m. This shows why that choice is correct.

The key idea is the projectile range without air resistance, given by R = v^2 sin(2θ) / g. With speed 20 m/s and launch angle 60°, compute sin(2θ) = sin(120°) = sin(60°) = √3/2 ≈ 0.866. Then R = (20^2) × 0.866 / 9.8 = 400 × 0.866 / 9.8 ≈ 346.4 / 9.8 ≈ 35.4 m. So the horizontal range is about 35 m, which matches the option described as approximately 35 m. This shows why that choice is correct.

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