UPSC NDA 2 2016 General Ability Solved Paper

Concept:Relative motion of two objects under same acceleration simplifies to constant relative velocity.Explanation:Let upward direction be positive.For ball A: initial velocity $u_A = +20\,\mathrm{m/s}$, acceleration $a_A = -g = -9.8\,\mathrm{m/s}^2$.For ball B: initial velocity $u_B = -20\,\mathrm{m/s}$ (downward), acceleration $a_B = -g$ (same direction).Relative acceleration $a_{AB} = a_A - a_B = 0$ because both have $-g$.Relative initial velocity $u_{AB} = u_A - u_B = 20 - (-20) = 40\,\mathrm{m/s}$.Initial relative separation $S = 40\,\mathrm{m}$ (height of B above A).Time to collide: $t = \frac{S}{u_{AB}} = \frac{40}{40} = 1\,\mathrm{s}$.Now find height from ground where collision occurs using ball A's motion:$s_A = u_A t + \frac{1}{2} a_A t^2 = (20)(1) + \frac{1}{2}(-9.8)(1)^2 = 20 - 4.9 = 15.1\,\mathrm{m}$.Thus collision happens after 1 second at 15.1 m above ground.Answer:The balls collide after 1 s at a height of 15.1 m from the ground.