Two particles are projected in air with speed v0, at angles θ1 and θ2 to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices a) angle of project: q1 > q2 b) time of flight: T1 > T2 c) horizontal range: R1 > R2 d) total energy: U1 > U2
Given that, Two particles are projected in air at speed u at angles θ1 and θ2 to the horizontal. Also given that height reached by the first particle is greater than that of the second ⟹H1>H2 Where, H1 is the height reached by first particle and H2 is the height reached by second particle. H1=2gu2sin2θ1 H2=2gu2sin2θ2 We have 2gu2sin2θ1>2gu2sin2θ2 ⟹sin2θ1>sin2θ2 ⟹(sinθ1−sinθ2)(sinθ1+sinθ2)>0 ⟹sinθ1>sinθ2 Let T1 and T2 be the time of flight of first and second particle T1=g2usinθ1 T2=g2usinθ2 if sinθ1>sinθ2 then, T1>T2
