question_answer

There are two angles of projection for which the horizontal range is the same? Prove that the sum of the maximum heights for these two angles does not depend upon the angle of projection.

Answer:

                If a projectile is projected with velocity u, making an angle \[\theta \] with the horizontal direction, then Horizontal range, \[R=\frac{{{u}^{2}}}{R}\sin 2\theta \] And Max. height, \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\]                 Case (i) if \[\theta =\alpha \], let \[R={{R}_{1}}\]and \[H={{H}_{1}}\]then              \[{{R}_{1}}=\frac{{{u}^{2}}\sin 2\alpha }{g}\]                                     .?. (i) and\[{{H}_{1}}=\frac{{{u}^{2}}}{2g}{{\sin }^{2}}\alpha \]                                                                                                                                ?? (ii) Case (ii) If \[\theta =\left( {{90}^{o}}-\alpha  \right)\], let \[R={{R}^{2}}\] and \[H={{H}_{2}}\], then \[{{R}^{2}}=\frac{{{u}^{2}}\sin 2\left( {{90}^{o}}-\alpha  \right)}{g}=\frac{{{u}^{2}}}{g}\sin \left( {{180}^{o}}-2\alpha  \right)=\frac{{{u}^{2}}}{g}\sin 2\alpha \]                                                               ?? (iii) \[{{H}_{2}}=\frac{{{u}^{2}}}{2g}{{\sin }^{2}}\left( {{90}^{o}}-\alpha  \right)=\frac{{{u}^{2}}}{2g}{{\cos }^{2}}\alpha \]                                                                                                ?? (iv) From (i) and (iii) ;\[{{R}_{1}}={{R}_{2}}\] \[{{H}_{1}}+{{H}_{2}}=\frac{{{u}^{2}}}{2g}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta  \right)=\frac{{{u}^{2}}}{2g}\]