A projectile is launched from ground level with initial velocity v0 = 40 m/s at an angle of 30 degrees above the horizontal. What is its horizontal range? (Take g = 10 m/s^2, sin(60°) = 0.866)
Answer and explanation
Correct answer: A. 138.6 meters
- Step 1: Range formula: R = (v0^2 sin(2theta)) / g. Step 2. Here theta = 30°, so 2theta = 60°. Step 3. R = (40^2 sin(60°)) / 10 = (1600 0.866) / 10 = 160 0.866 = 138.56 m = 138.6 m. Answer: 138.6 meters. Why not the others: 80 m uses sin(30°); 160 m neglects the angle; 69.3 m is half the range (maximum height horizontal position).
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