Answer:
Using, \[y=\,{{\upsilon
}_{0y}}t+\frac{1}{2}\,{{a}_{y}}{{t}^{2}},\,\] we get
\[0\,=({{\upsilon
}_{0}}\,\sin \,\beta )\,t-\frac{1}{2}\,(g\,\cos \theta ){{t}^{2}}\]
\[\therefore
\,\,t=\,\frac{2{{\upsilon }_{0}}\sin \beta }{g\,\cos \,\alpha }\]
Which is the time of
flight.
Now \[OQ\,={{\upsilon
}_{0}}\,\cos \,(\alpha +\,\beta )\,\times t\]
\[=\frac{{{\upsilon
}_{0}}\cos (\alpha +\beta )\,\times \,2\,{{\upsilon }_{0}}\sin \,\beta
}{g\,\cos \theta }\]
\[\therefore
\]\[R=\,OP\,\,=\,\frac{0Q}{\cos \alpha }\]
\[=\,\frac{2\upsilon
_{0}^{2}\,\sin \beta \,\,\cos \,(\alpha +\beta )}{g\,{{\cos }^{2}}\alpha }\]
R will be maximum at \[\beta
\] if
\[\frac{dR}{d\beta }=\,0\]
or \[\,\frac{2{{\upsilon }_{0}}}{g\,{{\cos }^{2}}\alpha }\,[\cos \,\beta \,\cos
\,\,(\alpha +\,\beta )\]
\[-\,\sin \,\beta \,\sin
\,(\alpha +\,\beta )]=0\]
or \[\tan \,(\alpha
+\,\beta )\,=\,\cot \,\beta \,=\,\tan \,({{90}^{o}}-\beta )\]
or \[\alpha +\,\beta
\,=\,{{90}^{o}}\,-\beta \]
or \[\beta
={{45}^{o}}\,-\frac{\alpha }{2}\,=\,\frac{\pi }{4}-\,\frac{\alpha }{2}.\]
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