Two inclined planes (AB) and (BC) are placed as shown in Fig. 2 (ABC).3 A particle is projected from the foot of the plane of angle ` alpha` along its line witn a velocity just sufficient to carry it to the top after which the particle slides down the other inclined plane. Fing the total time it will take to reach the pont (C ).

Answer

`t = sqrt(((2h)/(g)))[(1)/(sin alpha) + (1)/(sin beta)]`

Explanation

For a body sliding down an inclined plane, `a = g sin beta`
Using, `x = ut + (1)/(2) at^(2)`
`l = 0 + (1)/(2) (g sin beta) t_(2)^(2)`
`(h)/(sin beta) = (1)/(2) (g sin beta) t_(2)^(2), t_(2) sqrt((2h)/(g)) (1)/(sin beta)`
Similarly, `t_(1) = sqrt((2h)/(g) (1)/(sin alpha)`
`:. t = t_(1) + t_(2) = sqrt((2h)/(g)) [(1)/(sin alpha) + (1)/(sin beta)]`