Answer:
(i) When the angle of projection, \[\theta ={{45}^{\circ }}+\alpha ,\], the horizontal range is \[R=\frac{{{u}^{2}}\sin 2({{45}^{\circ }}+\alpha )}{g}=\frac{{{u}^{2}}\sin ({{90}^{\circ }}+2\alpha )}{g}\] \[=\frac{{{u}^{2}}\cos 2\alpha }{g}\] (ii) When the angle of projection, \[\theta ={{45}^{\circ }}-\alpha ,\], the horizontal range is \[R'=\frac{{{u}^{2}}\sin 2({{45}^{\circ }}-\alpha )}{g}=\frac{{{u}^{2}}\sin ({{90}^{\circ }}-2\alpha )}{g}\] \[=\frac{{{u}^{2}}\cos 2\alpha }{g}\] Clearly, \[R'=R\].
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