BHU PMT BHU PMT (Mains) Solved Paper-2005

  • question_answer

    Directions: In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:

    In the case of projectile motion of two projectiles, A and B are projected with the same speed at angles\[15{}^\circ \]and\[75{}^\circ \]respectively to the horizontal, then: (1) \[{{H}_{A}}<{{H}_{B}}\]                          (2) \[{{H}_{A}}>{{H}_{B}}\] (3) \[{{T}_{A}}>{{T}_{B}}\]                            (4) \[{{T}_{A}}<{{T}_{B}}\]

    A)  1 and 2 are correct

    B)  2 and 3 are correct

    C)  1 and 4 and correct

    D)  1, 2 and 4 are correct

    Correct Answer: C

    Solution :

                     When\[\alpha +\beta =90{}^\circ ,\]then horizontal range is same Maximum height\[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] and time of flight\[T=\frac{2u\,\,\sin \,\theta }{g}\] i.e.,   \[H\propto {{\sin }^{2}}\theta \]and \[T\propto \sin \theta \] \[\therefore \] \[{{H}_{B}}>{{H}_{A}}\]and\[{{T}_{B}}>{{T}_{A}}\]


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