JEE Main 201912 Apr, Morning shiftPhysicsMediumKinematics177× since 2002The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x² . If it were launched at an angle θ\thetaθ₀ with speed v₀ then (g = 10 ms^–2) :AOption A. θ0=cos−1(15){\theta _0} = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)θ0=cos−1(51) and v0=53{v_0} = {5 \over 3}v0=35 ms⁻¹BOption B. θ0=cos−1(25){\theta _0} = {\cos ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right)θ0=cos−1(52) and v0=35{v_0} = {3 \over 5}v0=53 ms⁻¹COption C. θ0=sin−1(25){\theta _0} = {\sin ^{ - 1}}\left( {{2 \over {\sqrt 5 }}} \right)θ0=sin−1(52) and v0=35{v_0} = {3 \over 5}v0=53 ms⁻¹DOption D. θ0=sin−1(15){\theta _0} = {\sin ^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)θ0=sin−1(51) and v0=53{v_0} = {5 \over 3}v0=35 ms⁻¹CheckSkip · show solution
