A particle is projected from level ground. Assuming projection point as origin, x-axis along horizontal and y-axis along vertically upwards. If particle moves in x-y plane and its path is given by y = ax - bx² where a, b are positive constants. Then match the physical quantities given in Column-I with the values given in Column-II. (g in Column II is acceleration due to gravity.)
Physics · Kinematics part-2 Projectile Motion · Motion in 2-D and ground to ground projectile
JEE Advanced

Question

A particle is projected from level ground. Assuming projection point as origin, x-axis along horizontal and y-axis along vertically upwards. If particle moves in x-y plane and its path is given by y = ax - bx² where a, b are positive constants. Then match the physical quantities given in Column-I with the values given in Column-II. (g in Column II is acceleration due to gravity.)

Physics · Kinematics part-2 Projectile Motion · Motion in 2-D and ground to ground projectile

G1 Physics · Accepted Answer

Compare y = ax - bx² with y = (uᵧ/uₓ)x - gx²/(2uₓ²). Thus a = uᵧ/uₓ and b = g/(2uₓ²), so horizontal speed uₓ = √[g/(2b)] = r. The horizontal range is R = a/b = p. Maximum height is the vertex value H = a²/(4b) = q. Time of flight is T = R/uₓ = √[2a²/(bg)] = s. Hence A-r, B-s, C-q, D-p, which is option C.