A plane surface is inclined making an angle $\beta$ above the horizon. A bullet is fired with the point of projection at the bottom of the inclined plane with a velocity u; then the maximum range is given by:

Suppose the particle be projected with a velocity u making an angle $\mathrm{\theta }$ with the horizontal. Suppose the particle strikes the inclined plane at A after time T. If x-axis is taken along the inclined plane and y-axis normal to the inclined plane. Then

${\mathrm{u}}_{\mathrm{x}} = \mathrm{u} \cos (\mathrm{\theta }-\mathrm{\beta }), {\mathrm{u}}_{\mathrm{y}} = \mathrm{u} \sin (\mathrm{\theta }-\mathrm{\beta })$

During the time of flight T, the displacement along y-axis is zero. Using equation

 ${\mathrm{s}}_{\mathrm{y}} = {\mathrm{u}}_{\mathrm{y}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{y}}{\mathrm{t}}^2$

$0 = \mathrm{u} \sin (\mathrm{\theta }-\mathrm{\beta })\mathrm{T}+\frac{1}{2}(-\mathrm{g} \cos \mathrm{\beta }){\mathrm{T}}^2$

$\mathrm{T} = \frac{2\mathrm{u} \sin (\mathrm{\theta }-\mathrm{\beta })}{\mathrm{g} \cos \mathrm{\beta }}$

Component of velocity along horizontal = u cos $\mathrm{\theta }$

Distance covered along horizontal

$\mathrm{OB} = (\mathrm{u} \cos \mathrm{\theta })\mathrm{T}$

        = $\mathrm{u} \cos \mathrm{\theta }\left[\frac{2\mathrm{u} \sin (\mathrm{\theta }-\mathrm{\beta })}{\mathrm{g} \cos \mathrm{\beta }}\right]$

        = $\frac{2{\mathrm{u}}^2\cos \mathrm{\theta } \sin (\mathrm{\theta }-\mathrm{\beta })}{\mathrm{g} \cos \mathrm{\beta }}$

Range R = OA

 $\mathrm{OA} = \frac{\mathrm{OB}}{\cos \mathrm{\beta }} =\frac{2{\mathrm{u}}^2 \cos \mathrm{\theta } \sin (\mathrm{\theta }-\mathrm{\beta })}{\mathrm{g} {\cos }^2\mathrm{\beta }}$

R = $\frac{2{\mathrm{u}}^2 \cos \mathrm{\theta } \sin (\mathrm{\theta }-\mathrm{\beta })}{\mathrm{g} {\cos }^2\mathrm{\beta }}$

   = $\left[\frac{{\mathrm{u}}^2}{\mathrm{g} {\cos }^2\mathrm{\beta }}\right][2 \sin (\mathrm{\theta }-\mathrm{\beta })\cos \mathrm{\theta }]$

 = $\frac{{\mathrm{u}}^2}{\mathrm{g} {\cos }^2 \mathrm{\beta }}\left[\sin \right(2\mathrm{\theta }-\mathrm{\beta })-\mathrm{sin\beta }]$

For a given u and $\mathrm{\beta }$, R is maximum, when sin($2\mathrm{\theta }-\mathrm{\beta }$) is maximum, i.e., when

   $\sin (2\mathrm{\theta }-\mathrm{\beta }) = 1 \mathrm{or} 2\mathrm{\theta } - \mathrm{\beta } = \frac{\mathrm{\pi }}{2}$

or $2\mathrm{\theta } = \mathrm{\beta }+\frac{\mathrm{\pi }}{2} \mathrm{or} \mathrm{\theta } = \frac{\mathrm{\beta }}{2}+\frac{\mathrm{\pi }}{2}$

Hence, the angle of projection $\mathrm{\theta }$, for maximum range up the inclined plane is given by

$\mathrm{\theta } = \frac{\mathrm{\beta }}{2}+\frac{\mathrm{\pi }}{4}$

${\mathrm{R}}_{\max } = \frac{{\mathrm{u}}^2}{\mathrm{g} {\cos }^2\mathrm{\beta }}[1-\sin \mathrm{\beta }]$

         = $\frac{{\mathrm{u}}^2(1-\sin \mathrm{\beta })}{\mathrm{g}(1-{\sin }^2\mathrm{\beta })} = \frac{{\mathrm{u}}^2}{\mathrm{g}(1+\sin \mathrm{\beta })}$