question_answer

Two bodies are thrown with the same initial velocity at angles \[\alpha \] and \[\left( \text{9}0{}^\circ -\alpha  \right)\] with the horizontal. What will be the ratio of (i) maximum heights attained by them and (ii) of horizontal ranges?

Answer:

                Horizontal range for angular projection of a projectile is given by \[R=\frac{{{u}^{2}}\sin 2\theta }{g}\] Case (i) When angle of projection, \[\theta ={{45}^{o}}+\alpha \], let \[R={{R}_{1}}\left( say \right)\], then                                          ??? (i) \[{{R}_{1}}=\frac{{{u}^{2}}\sin 2\left( {{45}^{\text{o}}}+\alpha  \right)}{g}=\frac{{{u}^{2}}}{g}\sin \left( {{90}^{\text{o}}}+2\alpha  \right)=\frac{{{u}^{2}}\cos 2\alpha }{g}\] Case (i) When angle of projection, \[\theta ={{45}^{o}}-\alpha \], let \[R={{R}_{2}}\left( say \right)\], then \[{{R}_{2}}=\frac{{{u}^{2}}\sin 2\left( {{45}^{o}}-\alpha  \right)}{g}=\frac{{{u}^{2}}}{g}\sin \left( {{90}^{\text{o}}}-2\alpha  \right)=\frac{{{u}^{2}}}{g}\cos 2\alpha \]                                                                             ??? (ii) From equation (i) and (ii), \[{{R}_{1}}={{R}_{2}}\]              Horizontal range, \[R=\frac{{{u}^{2}}}{g}\sin 2\text{ }\!\!\theta\!\!\text{ }\]    and    Max. height, \[H=\frac{{{u}^{2}}{{\sin }^{2}}\text{ }\!\!\theta\!\!\text{ }}{2g}\] Case (i) when \[\theta =\alpha \]; \[{{R}_{1}}=\frac{{{u}^{2}}}{g}\sin 2\alpha \] and \[{{H}_{1}}=\frac{{{u}^{2}}{{\sin }^{2}}\alpha }{2g}\] Case (ii) When \[\theta =\left( {{90}^{\text{o}}}-\alpha  \right)\]; \[{{R}_{2}}=\frac{{{u}^{2}}\sin 2\left( {{90}^{\text{o}}}-\alpha  \right)}{g}=\frac{{{u}^{2}}\sin 2\alpha }{g}\]and\[{{H}_{2}}=\frac{{{u}^{2}}{{\sin }^{2}}\left( {{90}^{\text{o}}}-\alpha  \right)}{g}=\frac{{{u}^{2}}{{\cos }^{2}}\alpha }{g}\] \[\therefore \]  \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{{{\sin }^{2}}\alpha }{{{\cos }^{2}}\alpha }={{\tan }^{2}}\alpha \]                and \[\frac{{{R}_{1}}}{{{R}_{2}}}=1\].