Consider a new cartesian coordinates in whcih $x$-axis is along inclined plane $O P$ and $O Y$ axis perpendicular to it as shown in figure. Consider the motion of projectile form OAP.


$$
\begin{aligned}
&a_y=-g \cos \alpha \\
&a_x=-g \sin \alpha \\
&\text { At } O \text { and } P, \\
&y=0 \\
&u_y=v_0 \sin \beta \\
&t=T
\end{aligned}
$$
where $T$ is time taken in reaching from point $O$ to point $P$.
(a) Considering motion along $O X$ axis :-
Given, $s_x=L_1, u_x=v_0 \cos \beta, a_x=-g \sin \alpha$
$$
\begin{aligned}
t=T=\frac{2 v_0 \sin \beta}{g \cos \alpha} \\
s_x=u_x t+\frac{1}{2} a_x t^2 \\
L=v_0 \cos \beta(T)+\frac{1}{2}(-g \sin \alpha) T^2 \\
L=& v_0 \cos \beta T-\frac{1}{2} g \sin \alpha T^2=T\left[v_0 \cos \beta-\frac{1}{2} g \sin \alpha T\right] \\
=& T\left[v_0 \cos \beta-\frac{1}{2} g \sin \alpha \times \frac{2 v_0 \sin \beta}{g \cos \alpha}\right] \\
=& \frac{2 v_0 \sin \beta}{g \cos \alpha}\left[v_0 \cos \beta-\frac{v_0 \sin \alpha \sin \beta}{\cos \alpha}\right] \\
=& \frac{2 v_0^2 \sin \beta}{g \cos { }^2 \alpha}[\cos \beta \cdot \cos \alpha-\sin \alpha \sin \beta]
\end{aligned}
$$
$$
L=\frac{2 v_0^2 \sin \beta}{g \cos ^2 \alpha} \cos (\alpha+\beta)
$$
(b) Considering motion along vertical upward direction perpendicular to $O X$ and motion of projectile along New $O Y$ axis :-
$$
s_y=0, u_{\mathrm{y}}=v_0 \sin \beta, a_y=-g \cos \alpha, t=T
$$
Applying equation,
$$
\begin{aligned}
&s_y=u_y t+\frac{1}{2} a_y t^2 \\
&0=v_0 \sin \beta T+\frac{1}{2}(-g \cos \alpha) T^2 \\
&0=v_0 \sin \beta T-\frac{g}{2} \cos \alpha(T)^2 \\
&T\left[v_0 \sin \beta-\frac{g \cos \alpha}{2} T\right]=0 \\
&\text { either } T=0 \text { or } v_0 \sin \beta-\frac{g T}{2} \cos \alpha=0 \\
&\frac{g T}{2} \cos \alpha=v_0 \sin \beta, T=\frac{2 v_0 \sin \beta}{g \cos \alpha}
\end{aligned}
$$
As $T=0$, corresponding to point $O$
Hence, $T=$ Time of flight $=\frac{2 v_0 \sin \beta}{g \cos \alpha}$
(c) For range $(L)$ will be maximum along new $O X$ axis from above relation.
Let $L$, it will be maximum when
$\sin \beta \cdot \cos (\alpha+\beta)$ should be maximum as $\alpha$ is constant angle of inclination of plane.
Consider
$$
\begin{aligned}
Z &=\sin \beta \cdot \cos (\alpha+\beta) \\
&=\sin \beta[\cos \alpha \cdot \cos \beta-\sin \alpha \cdot \sin \beta] \\
&=\frac{1}{2}\left[\cos \alpha \cdot \sin 2 \beta-2 \sin \alpha \cdot \sin ^2 \beta\right] \\
&=\frac{1}{2}[\sin 2 \beta \cdot \cos \alpha-\sin \alpha(1-\cos 2 \beta)] \\
z &=\frac{1}{2}[\sin 2 \beta \cdot \cos \alpha-\sin \alpha+\sin \alpha \cdot \cos 2 \beta] \\
&=\frac{1}{2}[\sin 2 \beta \cdot \cos \alpha+\cos 2 \beta \cdot \sin \alpha-\sin \alpha] \\
z &=\frac{1}{2}[\sin (2 \beta+\alpha)-\sin \alpha] \\
z \text { ismaximum, If } \\
\sin (2 \beta+\alpha)=1
\end{aligned}
$$
$$
\Rightarrow \quad \sin (2 \beta+\alpha)=\sin \frac{\pi}{2}
$$
So, $2 \beta+\alpha=\frac{\pi}{2}$
$$
\Rightarrow \beta=\left(\frac{\pi}{4}-\frac{\alpha}{2}\right) \text { radian }
$$