Question

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

Answer 1

Let ${\theta }_0$ be the angle of inclined plane with horizontal. Then for upward projection,
$R_{max}=\frac{u^2}{g(1+sin{\theta }_0)}=R_1$ ..(i)
For downward projection,
$R_{max}=\frac{u^2}{g(1-sin{\theta }_0)}=R_2$ ..(i)
For a projection on horizontal surface, we have
$R_{max}=\frac{u^2}{g}=R\left(say\right)$ ...(iii)
To establish a relation between $R,R_1,$ and $R_2$, we need to eliminate ${\theta }_0$. From (i) and (ii), we get
$\frac{2}{R}=\frac{1}{R_1}+\frac{1}{R_2}\Rightarrow R=\frac{2R_1{R}_2}{R_1+R_2}$