JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    The trajectory of a projectile near the surface of the earth is given as \[y=2x-9{{x}^{2}}.\]. If it were launched at an angle \[{{\theta }_{0}}\]with speed \[{{\text{v}}_{0}}\]then \[\left( g=10\text{ }m{{s}^{2}} \right)\]: [JEE Main Held on 12-4-2019 Morning]

    A) \[{{\theta }_{0}}={{\cos }^{-1}}\left( \frac{1}{\sqrt{5}} \right)\]and\[{{\text{v}}_{0}}=\frac{5}{3}m{{s}^{-1}}\]

    B) \[{{\theta }_{0}}={{\sin }^{-1}}\left( \frac{1}{\sqrt{5}} \right)\]and\[{{\text{v}}_{0}}=\frac{5}{3}m{{s}^{-1}}\]

    C) \[{{\theta }_{0}}={{\sin }^{-1}}\left( \frac{2}{\sqrt{5}} \right)\]and\[{{\text{v}}_{0}}=\frac{3}{5}m{{s}^{-1}}\]

    D) \[{{\theta }_{0}}={{\cos }^{-1}}\left( \frac{2}{\sqrt{5}} \right)\]and\[{{\text{v}}_{0}}=\frac{3}{5}m{{s}^{-1}}\]

    Correct Answer: A

    Solution :

    Equation of trajectory is given as \[y=2x9{{x}^{2}}\]                         ........ (1) Comparing with equation : \[y=x\tan \theta -\frac{g}{2{{u}^{2}}{{\cos }^{2}}\theta }.{{x}^{2}}\]         ........ (2) We get;\[\tan \theta =2\] \[\therefore \]    \[\] Also,\[\frac{g}{2{{u}^{2}}{{\cos }^{2}}\theta }=9\] \[\Rightarrow \frac{10}{2\times 9\times {{\left( \frac{1}{\sqrt{5}} \right)}^{2}}}={{u}^{2}}\]\[\Rightarrow {{u}^{2}}=\frac{25}{9}\] \[\Rightarrow \]


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