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. \( 15 \, \text{m} \)
• B. \( 20 \, \text{m} \)
• C. \( 30 \, \text{m} \)
• D. \( 45 \, \text{m}

Hint:

\textbf{Key Fact:} Maximum height depends on the vertical component of velocity (\( u \sin \theta \)) and decreases with higher gravity.

Answer 1

The Correct Option is A

• 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.