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)`)

Explanation

Two given planes are mutually perpendicular and the particle is projected perpendicularly from plane `OA`. It means `vec u` is parallel to plane `OB`.
At the instant of collision of the particle with `OB`, its velocity is perpendicular to `OB` or the velocity component parallel to `OB` is zero.
For considering the motion of particle parallel to plane `OB`,
`vec u = 10 sqrt(3) m s^-1`
Acceleration `= -g sin 60^@ = -5 sqrt(3) m s^-2`
`u = 0, t = ? S = ?`
Using `v = u + at, t = 2 s`
`s = ut + (1)/(2) at^2 or OQ = 10 sqrt(3)) m`
Now considering the motion of the particle normal to plane `OB`. initial velocity = 0, acceleration `= g cos 60^@ = 5 m s^-2`
`T = 2 s, v = ?, s = PO = ?`
Using `v = u + at, v = 10 m s^-1`
`s = ut + (1)/(2) at^2 or OP = 10 m`
`h = PO sin 30^@ = 10 xx sin 30^@ = 5 m`
The innclination of `vec u` with the vertical is `30^@`. Therefore, its vertical component is `u cos 30^@ = 15 m s^-1` (upward).
Considering vertically upward motion of the particle from `P`,
initial velocity `= 15 m s^-1`, acceleration `= g - 10 m s^-2, v = 0, S = H = ?`
Using, `v^2 = u^2 + 2 as, H = 11.25 m`
Therefore, maximum height reached by the particle above
`O = h + H = 16.25 m`
Distance, `PQ = sqrt((P Q)^2 + (O Q)^2) = sqrt((10)^2 + (10 sqrt(3))^2) = 20 m`.