Punjab Medical Punjab - MET Solved Paper-2011

  • question_answer Galileo writes that for angles of projection of a projectile at angles\[({{45}^{o}}+\theta )\]and\[({{45}^{o}}-\theta )\], the horizontal ranges described by the projectile are in the ratio

    A) \[2:1\]                                  

    B) \[1:2\]

    C) \[1:1\]                                  

    D) \[2:3\]

    Correct Answer: C

    Solution :

    Horizontal range\[R=\frac{{{u}^{2}}\sin 2\theta }{g}\Rightarrow R\propto \sin 2\theta \] \[\therefore \]  \[{{R}_{1}}\propto \sin 2({{45}^{o}}+\theta )\]                 \[{{R}_{1}}\propto \sin ({{90}^{o}}+2\,\,\theta )\] Similarly,\[{{R}_{2}}\propto \sin ({{90}^{o}}-2\,\,\theta )\] \[\because \]     \[\sin ({{90}^{o}}+2\theta )=\cos 2\theta \] \[\because \]     \[\sin ({{90}^{o}}-2\theta )=\cos 2\theta \] so,          \[\frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{1}\] or            \[{{R}_{1}}:{{R}_{2}}=1:1\]

adversite


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