A projectile is fired at an angle of 45 degrees to the horizontal. The ratio of its maximum height to its horizontal range is: MCQ with Answer and Explanation
A projectile is fired at an angle of 45 degrees to the horizontal. The ratio of its maximum height to its horizontal range is:
A. 1:2
B. 2:1
C. 4:1
D. 1:4
Answer: Option D
Solution (By JKExamLibrary)
The maximum height H = (u^2 * sin^2(theta)) / 2g. The horizontal range R = (u^2 * sin(2theta)) / g. For an angle of 45 degrees, sin(45) = 1/sqrt(2) and sin(90) = 1. H = (u^2 * 0.5) / 2g = u^2 / 4g. R = u^2 / g. Therefore, H/R = (u^2/4g) / (u^2/g) = 1/4. The ratio is 1:4.
Explanation:
Dispersion (splitting of white light into colors) occurs because refractive index n varies with wavelength (Cauchy's relation: n ∝ 1/λ² approximately). Since v = c/n, different colors travel at different speeds in the medium, causing different refraction angles (Snell's law). Frequency remains constant across media. Memory tip: 'Dispersion: n(λ) ⇒ v(λ) ⇒ different bending; violet bends most (highest n)'. This optics concept is frequently tested in competitive exams. Always distinguish cause (speed variation) from property (frequency/wavelength); competitive exams often test this nuanced understanding.
Explanation:
G = 6.67 × 10⁻¹¹ N m²/kg² in SI, and 6.67 × 10⁻⁸ dyne cm²/g² in CGS. Both are correct. Important constant, measured by Cavendish. g is acceleration due to gravity, different.
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