JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    If\[{{t}_{1}}\]and\[{{t}_{2}}\]are the times of flight of two particles having the same initial velocity u and range R on the horizontal, then\[t_{1}^{2}+t_{2}^{2}\]is equal to

    A) \[{{u}^{2}}/g\]                                  

    B) \[4{{u}^{2}}/{{g}^{2}}\]                

    C) \[{{u}^{2}}/2g\]         

    D)        1

    Correct Answer: B

    Solution :

    If two particles having same initial velocity\[u\]and range R, then their direction must be opposite i.e., the direction of projection of them are \[\alpha \] and\[90{}^\circ -\alpha \]. \[\therefore \]  \[{{t}_{1}}=\frac{2u\sin \alpha }{g}\] and       \[{{t}_{2}}=\frac{2u\sin ({{90}^{o}}-\alpha )}{g},\] \[\Rightarrow \]               \[{{t}_{2}}=\frac{2u\cos \alpha }{g}\] Now,     \[t_{1}^{2}+t_{2}^{2}=\frac{{{(2u\sin \alpha )}^{2}}}{{{g}^{2}}}+\frac{{{(2u\cos \alpha )}^{2}}}{{{g}^{2}}}\] \[=\frac{4{{u}^{2}}}{{{g}^{2}}}({{\sin }^{2}}\alpha +{{\cos }^{2}}ga)=\frac{4{{u}^{2}}}{{{g}^{2}}}\]

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