Mutually perpendicular x and y-axes are shown in the diagram.
The particle is projected from point O.
Let the time taken in reaching from point O to point P is T.

(b) Step 2: Considering motion along a vertically upward direction perpendicular to OX.

For the journey O to P.
$y=0,u_y=v_0\sin \beta ,a_y=-g\cos {\alpha }_{i}t=T$
Applying equation,

$\begin{array}{l}\mathrm{y}={\mathrm{u}}_{\mathrm{y}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{y}}{\mathrm{t}}^2\\ \Rightarrow 0={\mathrm{v}}_0\sin \mathrm{\beta T}+\frac{1}{2}(-\mathrm{gcos}\mathrm{\alpha }){\mathrm{T}}^2\\ \Rightarrow \mathrm{T}\left[{\mathrm{v}}_0\sin \mathrm{\beta }-\frac{\mathrm{gcos}\mathrm{\alpha }}{2}\mathrm{T}\right]=0\end{array}$

$\begin{array}{l}\Rightarrow \mathrm{T}=0,\mathrm{T}=\frac{2{\mathrm{v}}_0\sin \mathrm{\beta }}{\mathrm{gcos}\mathrm{\alpha }}\\ \text{ As }\mathrm{T}=0,\text{ corresponding to point }\mathrm{O}\\ \text{ Hence, }\mathrm{T}=\text{ Time of flight }=\frac{2{\mathrm{v}}_0\sin \mathrm{\beta }}{\mathrm{gcos}\mathrm{\alpha }}\end{array}$

(a) Step 3: Considering motion along OX.

$\begin{array}{l}\mathrm{x}={\mathrm{L}}_1, {\mathrm{u}}_{\mathrm{x}}={\mathrm{v}}_0\cos \mathrm{\beta }, {\mathrm{a}}_{\mathrm{x}}=-\mathrm{gsin}\mathrm{\alpha }\\ \mathrm{t} = \mathrm{T} = \frac{2{\mathrm{v}}_0\sin \mathrm{\beta }}{\mathrm{gcos}\mathrm{\alpha }}\\ \mathrm{x} = {\mathrm{u}}_{\mathrm{x}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{x}}{\mathrm{t}}^2\end{array}$

$\begin{array}{l}\Rightarrow \mathrm{L} = {\mathrm{v}}_0\cos \mathrm{\beta T} + \frac{1}{2}(-\mathrm{gsin}\mathrm{\alpha }){\mathrm{T}}^2\\ \Rightarrow \mathrm{L} = {\mathrm{v}}_0\cos \mathrm{\beta T} - \frac{1}{2}\mathrm{gsin}{\mathrm{\alpha T}}^2\\ = \mathrm{T}[{\mathrm{v}}_0\cos \mathrm{\beta }-\frac{1}{2}\mathrm{gsin}\mathrm{\alpha T}]\\ = \mathrm{T}[{\mathrm{v}}_0\cos \mathrm{\beta }-\frac{1}{2}\mathrm{gsin}\mathrm{\alpha }\times \frac{2{\mathrm{v}}_0\sin \mathrm{\beta }}{\mathrm{gcos}\mathrm{\alpha }}]\\ = \frac{2{\mathrm{v}}_0\sin \mathrm{\beta }}{\mathrm{gcos}\mathrm{\alpha }}[{\mathrm{v}}_0\cos \mathrm{\beta } - \frac{{\mathrm{v}}_0\sin \mathrm{\alpha sin}\mathrm{\beta }}{\cos \mathrm{\alpha }}]\\ =\frac{2{\mathrm{v}}_0^2\sin \mathrm{\beta }}{{\mathrm{gcos}}^2\mathrm{\alpha }}[\cos \mathrm{\beta }\cdot \cos \mathrm{\alpha }-\sin \mathrm{\alpha sin}\mathrm{\beta }]\\ \Rightarrow \mathrm{L} = \frac{2{\mathrm{v}}_0^2\sin \mathrm{\beta }}{{\mathrm{gcos}}^2\mathrm{\alpha }}\cos (\mathrm{\alpha }+\mathrm{\beta })\end{array}$

(c) Step 4: For range (L) to be maximum, $\sin \beta \cdot \cos (\alpha +\beta )$ should be maximum

$\begin{array}{l}\mathrm{Let}, \mathrm{Z}=\sin \mathrm{\beta }\cdot \cos (\mathrm{\alpha }+\mathrm{\beta })\\ =\sin \mathrm{\beta }\left[\cos \mathrm{\alpha }\cdot \cos \mathrm{\beta }-\sin \mathrm{\alpha }\cdot \sin \mathrm{\beta }\right]\\ =\frac{1}{2}\left[\cos \mathrm{\alpha }\cdot \sin 2\mathrm{\beta }-2\sin \mathrm{\alpha }\cdot {\sin }^2\mathrm{\beta }\right]\\ =\frac{1}{2}[\sin 2\mathrm{\beta }\cdot \cos \mathrm{\alpha }-\sin \mathrm{\alpha }(1-\cos 2\mathrm{\beta }\left)\right]\\ \Rightarrow \mathrm{z}=\frac{1}{2}[\sin 2\mathrm{\beta }\cdot \cos \mathrm{\alpha }-\sin \mathrm{\alpha }+\sin \mathrm{\alpha }\cdot \cos 2\mathrm{\beta }]\\ =\frac{1}{2}[\sin 2\mathrm{\beta }\cdot \cos \mathrm{\alpha }+\cos 2\mathrm{\beta }\cdot \sin \mathrm{\alpha }-\sin \mathrm{\alpha }]\\ =\frac{1}{2}[\sin (2\mathrm{\beta }+\mathrm{\alpha })-\sin \mathrm{\alpha }]\end{array}$

Step 5: For z to be maximum

$\begin{array}{l}\sin (2=\mathrm{\beta }+\mathrm{\alpha })=\text{ maximum }=1\\ \Rightarrow 2\mathrm{\beta }+\mathrm{\alpha }=\frac{\mathrm{\pi }}{2}\text{ or, }\mathrm{\beta }=\frac{\mathrm{\pi }}{4}-\frac{\mathrm{\alpha }}{2}\end{array}$