Practice question

Given: A stone is dropped from a height of 80 m while another is thrown upwards at 20 m/s from the ground at the same instant. When do they meet? (Take g = 10 m/s² ) These values define the system as per NCERT data. Formula: Stone 1: y_1 = 80 - 1/2 · 10 · t² = 80 - 5 t². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: Stone 2: y_2 = 20 t - 1/2 · 10 · t² = 20 t - 5 t² . Total height = 80 m, so y_1 + y_2 = 80 . Substitute: (80 - 5 t²) + (20 t - 5 t²) = 80 Rightarrow 80 + 20 t - 10 t² = 80 Rightarrow 20 t - 10 t² = 0 Rightarrow t (2 - t) = 0 . t = 2 s (discard t = 0 ). 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.