A body is projected at 60o with ground. It covers a horizontal distance of 100 m. If the same body is projected at 600 with vertical with same velocity, the new range is

Range, R = u2 sin 2θgFor same v, R ∝ 2θFor θ = 600 with horizontal, R is given as 100 m when R ∝ sin 1200--------(i)For  θ = 600 with vertical        R' ∝ sin 2 (900-600)i.e., R' ∝ sin 600---------------(ii)Using (i) and (ii)R'R = sin 600sin 1200 = 11i.e., R' = R = 100 m