For a given velocity of projection from a point on the inclined plane, the maximum range down the plane is three times the maximum range up the incline. Then find the angle of inclination of the inclined plane.

Answer

3

Explanation

3
`(u^(2))/(g[1-sinbeta])=(3u^(2))/(g[1+sin beta])implies1+sinbeta=3-3sinbetaimpliessinbeta=1/2impliesbeta=30^(@)`