question_answer

  Two bodies are thrown with the same initial velocity at angles \[\alpha \] and \[\left( \mathbf{90}{}^\circ \text{ }-\text{ }\alpha  \right)\] with the horizontal. What will be the ratio of (i) maximum heights attained by them and (ii) their horizontal ranges?

Answer:

                (i) When the angle of projection\[\alpha \], the maximum height is \[{{H}_{1}}=\frac{{{u}^{2}}{{\sin }^{2}}\alpha }{2g}\] When the angle of projection is\[\left( \text{9}0{}^\circ \text{ }-\text{ }\alpha  \right)\], the maximum height is \[{{H}_{2}}=\frac{{{u}^{2}}{{\sin }^{2}}({{90}^{\circ }}-\alpha )}{2g}=\frac{{{u}^{2}}{{\cos }^{2}}\alpha }{2g}\] \[\therefore \]  \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{{{\sin }^{2}}\alpha }{{{\cos }^{2}}\alpha }={{\tan }^{2}}\alpha =\mathbf{ta}{{\mathbf{n}}^{\mathbf{2}}}\mathbf{\alpha :1}\] (ii) When the angle of projection is a, the horizontal range is \[{{R}_{1}}=\frac{{{u}^{2}}\sin 2\alpha }{g}\] When the angle of projection is\[\left( \text{9}0{}^\circ -\alpha  \right)\], the horizontal range is \[{{R}_{2}}=\frac{{{u}^{2}}\sin 2({{90}^{\circ }}-\alpha )}{g}=\frac{{{u}^{2}}\sin ({{180}^{\circ }}-2\alpha )}{g}\] \[=\frac{{{u}^{2}}\sin 2\alpha }{g}\] \[\therefore \]  \[{{R}_{1}}:{{R}_{2}}=\mathbf{1:1}\]