A projectile has same range R for the two angles of projection. If T1 and T2 are the times of flight in two cases, then:
T1T2∝R
T1T2∝R2
T1T2∝1R
T1T2∝1R2
Range is same for angle of projection θ and (90−θ)T1T2=2usinθg⋅2usin(90−θ)gT1T2=4u2sinθcosθg22g⋅u2sinθg=2Rg⇒T1T2∝RHence correct choice is (1).