Question

A particle is projected with a certain velocity at an angle alpha above the horizontal from the foot of inclined plane of inclination 30.if the particle strikes the plane normally then alpha is equal to 1)30 + tan inverse of square root of 3 by 2 2)45 3)60 4)30 - tan inverse 2 × square root of 3

Answer 1

$$\begin{gathered} Dearstudent, \\ timeofflightoftheparticle=2u\sin (\alpha -{30}^o) \\ NowcomponentofvelocityalongtheplanebecomeszeroatpointB. \\ 0=u\cos (\alpha -30°)-g\sin 30°T \\ =u\cos (\alpha -30°)=(g\sin 30°)2u\sin (\alpha -30°) \\ \Rightarrow \alpha =30°+({\tan }^{-1})\left(\sqrt{\frac{3}{2}}\right) \\ Regards \end{gathered}$$