Practice question

Given: A stone is dropped from a cliff 45 m high with an initial horizontal speed of 15 m/s. What is its speed on hitting the ground? (Take g = 10 m/s²) These values define the system as per NCERT data. Formula: Horizontal velocity: v_x = 15 m/s (constant). This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Vertical velocity: v_y = √(2gh) = √(2 × 10 × 45) = √900 = 30 m/s. Speed v = √(v_x² + v_y²) = √(15² + 30²) = √(225 + 900) = √1125 ≈ 33.54 m/s. 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.