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Sahay Sir > Question Answers > JEE Mains Physics 2002-2019 Solved Video Solutions > Chapter 2 - Kinematics > The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x^2 . If it were launched at an angle θ0 with speed v0 then g = 10 ms^–2

The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x^2 . If it were launched at an angle θ0 with speed v0 then g = 10 ms^–2

The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x^2 . If it were launched at an angle θ0 with speed v0 then g = 10 ms^–2