Practice question

Given: A projectile is launched at 20 m/s at 53° . What is its maximum height? (Take g = 10 m/s², sin 53° = 0.8 ) These values define the system as per NCERT data. Formula: Maximum height h_m = (v_0 sin θ_0)²/2g. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Given: v_0 = 20 m/s, sin 53° = 0.8, g = 10 m/s² . h_m = (20 × 0.8)²/2 × 10 = 16²/20 = 256/20 = 12.8 m . Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.