Two inclined planes OA and OB having inclinations `30^@` and `60^@` with the horizontal respectively intersect each other at O, as shown in figure. A particle is projected from point P with velocity `u=10 sqrt(3) m//s` along a direction perpendicular to plane OA. If the particle strikes plane OB perpendicular at Q. Calculate.

(a) time of flight,
(b) velocity with which the particle strikes the plane OB,
(c) height h of point P from point O,
(d) distance PQ. (Take `g=10m//s^(2)`)

Answer

(i) 2s (ii) `10 ms^(-1)` (iii) 5m (iv) `16.25 m` (v) 20 m

Explanation

(i) `v(t)=(u -g cos 30^(@) t) hat(i) -g sin theta t hat(j)`
From given situation
`u -g cos 30^(@) t = 0`
`t =2 sec`
(ii) Velocity `u_(x) = 0,a_(x) = g cos 30^(@) = (g)/(2)`
`:. v_(x) =0 +(g)/(2) xx 2 = 10 m//s`
(iii) Distance `PO =`
`10 sqrt(3) cos 90^(@) xx t +(1)/(2) xxg sin 30^(@) xx (2)^(2)`
`PO = 10 m :. h = 10 sin 30^(@) = 5 m`
(iv) Maximum height `= h + (u(sin 60^(@))^(2))/(2g)`
`= 5 +((10 sqrt(3) xx (sqrt(3))/(2)))/(20) =16.25 m`
(v) Distance PQ
`OQ = ((10 sqrt(3))^(2))/(2g cos 30^(@))`
`OQ =1 0 sqrt(3)`
`:. PQ = sqrt((PO)^(2) +(O)^(2))`
`= sqrt(10^(2)+(10 sqrt(3))^(2))= 20m`.