Uttarakhand PMT Uttarakhand PMT Solved Paper-2007

  • question_answer
    The angle of projection for which the horizontal range and the maximum height of the projectile are equal is

    A)  \[45{}^\circ \]           

    B)  \[\text{ }\!\!\theta\!\!\text{ =ta}{{\text{n}}^{-1}}(4)\]

    C)  \[\text{ }\!\!\theta\!\!\text{ =ta}{{\text{n}}^{-1}}(0.25)\]

    D)  none of these

    Correct Answer: B

    Solution :

     Given, horizontal range = maximum height \[\Rightarrow \] \[\frac{{{u}^{2}}\sin 2\theta }{g}=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] \[\Rightarrow \] \[\frac{2{{u}^{2}}\sin \theta \cos \theta }{g}=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] \[\Rightarrow \] \[2\cos \theta =\frac{\sin \theta }{2}\] \[\Rightarrow \] \[\frac{\sin \theta }{\cos \theta }=4\] \[\Rightarrow \] \[\tan \theta =4\] \[\therefore \] \[\theta ={{\tan }^{-1}}(4)\]


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