See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

(for Q. 18-19) 2 2u sin cos R g    Impulse of normal on brick at landing = mu sin  Impulse due to friction = mu sin  = mu cos   mvx x v u(cos sin )    Horizontal distance covered = 2 x v R 2 g   = 2 2 2 2u sin cos u (cos sin ) g 2 g       2 2 u d (cos sin ) 2 g     For maximum distance  = tan  But this is valid only for vX > 0  cos    sin  > 0 1 tan E + dE E dx x AITS-FT-VII (Paper-2)-PCM(Sol.)-JEE(Advanced)/22 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 5 Chemistry PART – II Section – A