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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2011

  • question_answer
    Two projectiles A and B thrown with speeds in the ratio 1 : \[\sqrt{2}\] acquired the same heights. If A is thrown at an angle of \[45{}^\circ \] with the horizontal, the angle of projection of B will be

    A) \[0{}^\circ \]                      

    B)        \[60{}^\circ \]                   

    C) \[30{}^\circ \]                   

    D)        \[45{}^\circ \]

    E) \[15{}^\circ \]

    Correct Answer: C

    Solution :

    Given condition \[{{h}_{1}}={{h}_{2}}\] \[u_{1}^{2}{{\sin }^{2}}{{45}^{o}}=u_{2}^{2}{{\sin }^{2}}\theta \]                 \[{{\sin }^{2}}\theta =\frac{u_{1}^{2}}{u_{2}^{2}}{{\sin }^{2}}{{45}^{o}}\]                 \[=\frac{1}{2}.\frac{1}{2}=\frac{1}{4}\] \[\sin \theta =\frac{1}{2}\] \[\Rightarrow \]               \[\theta ={{30}^{o}}\]


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