$A$ particle is projected in air at an angle $\beta$ to a surface which itself is inclined at an angle $\alpha$ to the horizontal (figure).
$(a)$ Find an expression for the range on the plane surface (distance on the plane from the point of projection at which the particle will hit the surface).
$(b)$ Find the time of flight.
$(c)$ Find the angle $\beta$ at which the range will be maximum.

(N/A) Consider the adjacent diagram.
Let the coordinate system be set such that the $X$-axis is along the inclined plane and the $Y$-axis is perpendicular to it.
Initial velocity components: $U_x = v_0 \cos \beta$,$U_y = v_0 \sin \beta$.
Acceleration components: $a_x = -g \sin \alpha$,$a_y = -g \cos \alpha$.
$(b)$ Time of flight $(T)$:
At the point of impact $P$,the displacement along the $Y$-axis is $y = 0$.
Using $y = U_y T + \frac{1}{2} a_y T^2$:
$0 = (v_0 \sin \beta) T - \frac{1}{2} (g \cos \alpha) T^2$
$T = \frac{2 v_0 \sin \beta}{g \cos \alpha}$.
$(a)$ Range $(R)$:
The range is the displacement along the $X$-axis at time $T$.
$R = U_x T + \frac{1}{2} a_x T^2$
$R = (v_0 \cos \beta) \left( \frac{2 v_0 \sin \beta}{g \cos \alpha} \right) - \frac{1}{2} (g \sin \alpha) \left( \frac{2 v_0 \sin \beta}{g \cos \alpha} \right)^2$
$R = \frac{2 v_0^2 \sin \beta \cos \beta}{g \cos \alpha} - \frac{2 v_0^2 \sin^2 \beta \sin \alpha}{g \cos^2 \alpha}$
$R = \frac{2 v_0^2 \sin \beta}{g \cos^2 \alpha} [\cos \beta \cos \alpha - \sin \beta \sin \alpha]$
$R = \frac{2 v_0^2 \sin \beta \cos(\alpha + \beta)}{g \cos^2 \alpha}$.
$(c)$ Maximum range:
For maximum range,$\frac{dR}{d\beta} = 0$.
Using the identity $2 \sin \beta \cos(\alpha + \beta) = \sin(2\beta + \alpha) - \sin \alpha$,we get:
$R = \frac{v_0^2}{g \cos^2 \alpha} [\sin(2\beta + \alpha) - \sin \alpha]$.
For $R$ to be maximum,$\sin(2\beta + \alpha) = 1$,so $2\beta + \alpha = 90^\circ$.
$\beta = 45^\circ - \frac{\alpha}{2}$.