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The trajectory of a projectile near the surface of the earth is given as \( y=2 x-9 x^{2} \), If it were launched at an angle \( \theta_{0} \) with speed \( v_{0} \) then \( \left(g=10 m^{-2}\right) \)
[JEE (Main) 2019]
(a) \( \theta_{0}=\cos ^{-1}\left(\frac{1}{\sqrt{5}}\right) \) and \( v_{0}=\frac{5}{3} m s^{-1} \)
(c) \( \theta_{0}=\cos ^{-1}\left(\frac{2}{\sqrt{5}}\right) \) and \( v_{0}=\frac{3}{5} m s^{-1} \)
(d) \( \theta_{0}=\sin ^{-1}\left(\frac{2}{\sqrt{5}}\right) \) and \( v_{0}=\frac{3}{5} m s^{-1} \)
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