Practice question

Given: Two stones are dropped from a height of 200 m, with a 2.5 s interval. What is their separation when the second stone has fallen for 3 s ? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: First stone (after 5.5 s): y_1 = 1/2 · 10 · (5.5)² = 151.25 m. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Second stone (after 3 s): y_2 = 1/2 · 10 · (3)² = 45 m . Separation = 200 - 151.25 - 45 = 3.75 m (but both falling, so 151.25 - 45 = 106.25 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.