Practice question

Given: A ball is thrown upwards at 20 m/s from a 50 m tower. What is its speed when it passes 20 m below the tower’s top? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Displacement y = -20 m, v² = (20)² + 2 · 10 · 20 = 400 + 400 = 800. This is standard NCERT relation. Substitution & Calculation: v = sqrt800 approx 28.28 m/s . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.