2. Questions Bank
  3. JEE Main & Advanced
  4. Physics

JEE Main & Advanced Physics Question Bank Oblique Projectile Motion

  • question_answer
    A stone projected with a velocity u at an angle q with the horizontal reaches maximum height \[{{H}_{1}}\].  When it is projected with velocity u at an angle \[\left( \frac{\pi }{2}-\theta  \right)\] with the horizontal, it reaches maximum height \[{{H}_{2}}\].  The relation between the horizontal range R of the projectile, \[{{H}_{1}}\]and \[{{H}_{2}}\] is                                             [EAMCET 2000]

    A)             \[R=4\sqrt{{{H}_{1}}{{H}_{2}}}\]          

    B)             \[R=4({{H}_{1}}-{{H}_{2}})\]

    C)             \[R=4({{H}_{1}}+{{H}_{2}})\]

    D)               \[R=\frac{{{H}_{1}}^{2}}{{{H}_{2}}^{2}}\]

    Correct Answer: A

    Solution :

                    \[{{H}_{1}}=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] and \[{{H}_{2}}=\frac{{{u}^{2}}{{\sin }^{2}}(90-\theta )}{2g}=\frac{{{u}^{2}}{{\cos }^{2}}\theta }{2g}\]             \[{{H}_{1}}{{H}_{2}}=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\times \frac{{{u}^{2}}{{\cos }^{2}}\theta }{2g}=\frac{{{({{u}^{2}}\sin 2\theta )}^{2}}}{16{{g}^{2}}}=\frac{{{R}^{2}}}{16}\]             \\[R=4\sqrt{{{H}_{1}}{{H}_{2}}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner