An inclined plane makes an angle θ = 45° with horizontal. A stone is projected normally from the inclined plane, with speed u m/s at t = 0 sec. x and y axis are drawn from point of projection along and normal to inclined plane as shown. The length of incline is sufficient for stone to land on it and neglect air friction. Match the statements given in Column-I with the results in Column-II. (g in Column-II is acceleration due to gravity.)

Question

G1 Physics · Accepted Answer

Take x along the 45° incline and y normal to it. The stone is projected normally, so uₓ = 0, uᵧ = u, while gravity components are aₓ = g/√2 and aᵧ = -g/√2. A: vᵧ = 0 gives t = √2u/g = r. B: vₓ = vᵧ gives t = u/(√2g) = s. D: y = 0 again gives time of flight T = 2√2u/g = p. C: x ∝ t², so half the total range occurs at t = T/√2 = 2u/g = q. Thus A-r, B-s, C-q, D-p, which is option B.