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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 19

  • question_answer
    A projectile is thrown in the upward direction making an angle of \[60{}^\circ \] with the horizontal direction with a velocity of \[147\,m{{s}^{-1}}\]. Then the time after which its inclination with the horizontal is \[45{}^\circ \] is:

    A) \[15\,s\]            

    B)        \[10.98\,s\] 

    C) \[5.49\,s\]          

    D)        \[2.745\,s\]

    Correct Answer: C

    Solution :

    \[u=147m/s,\] \[\theta =60{}^\circ \] Vertically, \[v\sin \phi =u\sin \theta -gt\] \[\Rightarrow \,\,9.8t=u\sin 60{}^\circ -v\sin 45{}^\circ \] \[=\frac{u\sqrt{3}}{2}-\frac{v}{\sqrt{2}}\]                      ?.(i) Horizontally, \[u\cos \theta =v\cos \phi \] \[\Rightarrow \,\,u\cos 60{}^\circ =v\cos 45{}^\circ \] or         \[v=\frac{u}{\sqrt{2}}\]                         ?..(ii) \[\therefore \,\,t=\frac{1}{2(9.8)}(\sqrt{3}-1)147=5.49\,\sec \]


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