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NEET Sample Paper NEET Sample Test Paper-17

  • question_answer
    A particle is projected at an angle of \[60{}^\circ \] above the horizontal with a speed of 10 m/s. After some time the direction of its velocity makes an angle of \[30{}^\circ \] above the horizontal. The speed of the particle at this instant is

    A) \[\frac{5}{\sqrt{3}}m/s\]                   

    B) \[5\sqrt{3}\,m/s\]                    

    C) \[5\,m/s\]                     

    D) \[\frac{10}{\sqrt{3}}\,m/s\] 

    Correct Answer: D

    Solution :

    Let v be the velocity of projectile at this instant. Horizontal component of velocity remains unchanged. Therefore, \[v\cos {{30}^{o}}=10\cos {{60}^{o}}\] or \[v\frac{\sqrt{3}}{2}=\frac{10}{2}\] \[\therefore \]    \[v=\frac{10}{\sqrt{3}}m/s\] Hence, the correction option is [d].


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