Question:
A particle is projected at an angle of \( 60^\circ \) to the horizontal with an initial speed of \( 20 \, \text{m/s} \). If \( g = 10 \, \text{m/s}^2 \), what is the maximum height reached by the particle?
A particle is projected at an angle of \( 60^\circ \) to the horizontal with an initial speed of \( 20 \, \text{m/s} \). If \( g = 10 \, \text{m/s}^2 \), what is the maximum height reached by the particle?
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\textbf{Key Fact:} Maximum height depends on the vertical component of velocity (\( u \sin \theta \)) and decreases with higher gravity.
Updated On: May 26, 2025
- \( 15 \, \text{m} \)
- \( 20 \, \text{m} \)
- \( 30 \, \text{m} \)
- \( 45 \, \text{m}
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The Correct Option is A
Solution and Explanation
- Formula for Maximum Height: For a projectile, the maximum height H is given by H = (u² sin² θ) / (2g), where u is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.
- Given Values: u = 20 m/s, θ = 60°, g = 10 m/s², and sin 60° = √3/2 ≈ 0.866.
- Calculate sin² θ: sin² 60° = (√3/2)² = 3/4 = 0.75.
- Compute Maximum Height:
H = (20)² × (3/4) ÷ (2 × 10) = (400 × 0.75) ÷ 20 = 300 ÷ 20 = 15 m. - Conclusion: The maximum height is (1) 15 m.
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