A body has maximum range $R_1$ when projected up the inclined plane. The same body when projected down the inclined plane. It has maximum range $R_2$. Find its maximum horizontal range. Assume the equal speed of projection in each case and the body is projected onto the greatest slope.

For upward projection, $R_{max}=\frac{V_0^2}{g(1-\sin \beta )}=R_1$   ………..(i)

For downward projection,  $R_{max}=\frac{v_0^2}{g(1+\sin \beta )}=R_2$   ……….(ii)

For a projection of horizontal surface substituting $\beta =0$, Then, we have, $R_{max}=\frac{V_0^2}{g}=R\left(say\right)$  ……..(iii)

To establish a relation between $R,R_1$ and $R_2$ we need to eliminate $\sin \beta$.

Adding $\frac{1}{R_2}$ form eq.(i)  with $\frac{1}{R_2}$ from eq.(ii) we have $\frac{2}{R}=\frac{1}{R_1}+\frac{1}{R_2}$  Then, 

Ans- $R=\frac{2R_1{R}_2}{R_1+R_2}$