Question

A particle is projected from the bottom of an inclined plane of inclination ${30}^{\circ }.$ At what angle $\alpha$ (from the horizontal) should the particle be projected to get the maximum range on the inclined plane.

Answer 1

Range on the inclined plane $R=\frac{ucos(30+\beta )T}{cos\left({30}^o\right)}$ where $T=\frac{2usin\beta }{gcos\left({30}^o\right)}$

For range to be maximum, ${30}^o+2\beta ={90}^o$

$⟹$ $\beta ={30}^o$

$⟹$ $\alpha ={30}^o+\beta ={60}^o$