Manipal Engineering Manipal Engineering Solved Paper-2015

  • question_answer
    A projectile is thrown in upward direction making an angle of 60° with the horizontal direction with a velocity of 150 \[2x+y=0\] Then, the time after which its inclination with horizontal is 45°, is

    A) \[\sin A+\cos A=m\]                      

    B) \[{{\sin }^{3}}A+{{\cos }^{3}}A=n,\]

    C) \[~{{m}^{3}}-3m+n=0\]                

    D) \[{{n}^{3}}-3n+2m=0\]

    Correct Answer: C

    Solution :

     At the two points of the trajectory during projectile motion, the horizontal component of velocity is same. Then, \[IV<III<II<I\] Initially,  \[I<II<III<IV\] Finally,   \[NO_{3}^{-}\] But\[C{{l}^{-}}\] \[{{l}^{-}}\]\[B{{r}^{-}}\]\[HOCl\] \[HCl{{O}_{2}}\]\[HCl{{O}_{3}}\]


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