Given data:
[tex]\begin{gathered} \cdot\text{The angle of initial velocity above the inclined plane is }\phi \\ \cdot\text{ The angle of the inclined plane with the horizontal is }\theta \end{gathered}[/tex]Solution:
(a). The angel of the initial velocity with respect to the horizontal surface is,
[tex]\alpha=\theta+\phi[/tex]By the kinematic equation, the horizontal range in terms of initial velocity is,
[tex]R=v_{\circ}\cos (\alpha)t+\frac{1}{2}a_xt^2[/tex]where a_x is the acceleration of the ball along the horizontal direction,
As no force on the ball is acting along the horizontal direction, thus, the acceleration of the ball is zero.
[tex]R=v_o\cos (\alpha)t[/tex]By the kinematic equation along the vertical direction, the acceleration due to gravity in terms of the velocity of the ball is,
[tex]\begin{gathered} g=\frac{v_o\sin(\alpha)}{t_1} \\ t_1=\frac{v_o\sin (\alpha)}{g} \end{gathered}[/tex]where t_1 is the time taken to reach the maximum height,
The total time of the projectile motion in terms of t_1 is,
[tex]\begin{gathered} t=2t_1 \\ t=\frac{2v_o\sin (\alpha)}{g} \end{gathered}[/tex]Thus, the horizontal range becomes,
[tex]\begin{gathered} R=\frac{2v^2_o\cos (\alpha)\sin (\alpha)}{g} \\ R=\frac{v^2_o(2\sin (\alpha)\cos (\alpha))}{g} \\ R=\frac{v^2_o\sin (2\alpha)}{g} \end{gathered}[/tex]Thus, the horizontal range of the ball in terms of the initial velocity and angle above the horizontal surface is,
[tex]\begin{gathered} R=\frac{v^2_o\sin (2\alpha)}{g} \\ R=\frac{v^2_o\sin (2\theta+2\phi)}{g} \end{gathered}[/tex]where g is the acceleration due to gravity,
The distance covered by the ball along the inclined plane is,
[tex]\begin{gathered} R_1=R\cos (\theta) \\ R_1=\frac{v^2_o\sin(2\theta+2\phi)\cos(\theta)}{g} \end{gathered}[/tex]Thus, the distance measured along the incline from the launch point is,
[tex]\frac{v^2_o\sin(2\theta+2\phi)\cos(\theta)}{g}[/tex](b). The maximum range of the ball can be obtained at the maximum value of sine in the horizontal range term.
[tex]\frac{d}{d\phi}(R)=0[/tex]Substituting the known values,
[tex]\begin{gathered} \frac{d}{d\phi}(\frac{v^2_o\sin(2\theta+2\phi)\cos(\theta)}{g})=0 \\ \frac{v^2_o\cos(\theta)}{g}\times\frac{d}{d\phi}(\sin (2\theta+2\phi))=0 \\ \frac{d}{d\phi}(\sin (2\theta+2\phi))=0 \\ As\text{ }\frac{v^2_o\cos(\theta)}{g}\text{ term cannot be equal to zero.} \end{gathered}[/tex]By solving,
[tex]\begin{gathered} \cos (2\theta+2\phi)=0 \\ 2\theta+2\phi=\frac{\pi}{2}^{} \\ 2\phi=\frac{\pi}{2}-2\theta \\ \phi=\frac{\pi}{4}-\theta \end{gathered}[/tex]Thus, the maximum range of ball is obtained when the value of phi is,
[tex]\phi=\frac{\pi}{4}-\theta[/tex]