NEET Sample Paper NEET Sample Test Paper-19

  • question_answer A particle is projected from the ground with an initial speed of v at an angle\[\theta \] with horizontal. The magnitude of the average velocity of the particle between its point of projection and highest point of trajectory is:

    A)  \[\frac{v}{2}\sqrt{1+2{{\cos }^{2}}\theta }\]      

    B) \[\frac{v}{2}\sqrt{1+2{{\cos }^{2}}\theta }\]

    C) \[\frac{v}{2}\sqrt{1+3{{\cos }^{2}}\theta }\]       

    D) \[v\cos \theta \]  

    Correct Answer: C

    Solution :

    Initial velocity \[=v\cos \theta \hat{i}+v\sin \theta \hat{j}\] Final velocity \[=v\cos \theta \,\hat{i}\] The\[x\]component of the velocity remains \[v\cos \theta ,\]so its average value is also \[v\cos \theta ,\]whereas the y component of the velocity changes from \[v\sin \theta \]to, zero, so its average value is\[\frac{v\sin \theta }{2}.\] \[\therefore \]Average velocity\[=v\cos \theta \,\hat{i}+\frac{v\sin \theta \hat{j}}{2}\] Magnitude of average velocity \[=\frac{1}{2}\sqrt{4{{v}^{2}}{{\cos }^{2}}\theta +{{v}^{2}}{{\sin }^{2}}\theta }\] \[=\frac{1}{2}\sqrt{{{v}^{2}}+3{{v}^{2}}{{\cos }^{2}}\theta }=\frac{v}{2}\sqrt{1+3{{\cos }^{2}}\theta }\] Hence, the correction option is [c].


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