question_answer

Prove that the horizontal range is same when angle of projection is (i) greater than \[\mathbf{45}{}^\circ \] by certain value and (ii) less than \[\mathbf{45}{}^\circ \] by the same value.

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,   \[R'=R\].