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 \[\frac{{{R}_{1}}}{{{R}_{2}}}=1\].
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