Projectile Range
A javelin is launched at 40 m/s at 45 degrees to the horizontal across a level field, with g equal to 10 m/s², what is its horizontal range?
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About This Question
- Subject
- physics
- Chapter
- kinematics
- Topic
- projectile range
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
160 m
Per NCERT Class 11, Chapter 4 (Motion in a Plane), the horizontal range of a projectile on level ground is R = (u² sin2θ) / g, which combines the launch speed, the launch angle, and gravity. With u = 40 m/s, θ = 45°, sin2θ = sin90° = 1, so R = (40² × 1) / 10 = 1600/10 = 160 m. The option 80 m results from forgetting to square the velocity or halving the result. The option 320 m doubles the range by mistakenly using 2u² in the numerator. The option 120 m has no consistent derivation and underestimates the range. Plausibility check: 45° is the angle of maximum range for a given speed, and the time of flight is 2u sinθ/g = 2(40)(0.707)/10 ≈ 5.66 s; multiplying by the horizontal velocity u cosθ ≈ 28.3 m/s gives about 160 m, confirming the standard range formula result.
This medium difficulty physics question is from the chapter kinematics, covering the topic of projectile range. It appeared in the 2025 exam.
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