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

(a) `2sec`, (b) `10ms^(-1)`,(c ) `5m`, (d) `16.25m`, (e) `20m`

Explanation

(a) `u_(x) = 0, u_(y) = 10 sqrt(3) m//s`,
`v_(x) = 0, a_(x) = g cos 60^(@), a_(y) =- g sin 60^(@)`
`v_(y) = u_(y) +a_(y)t rArr 0 = 10 sqrt(3) - g sin 60^(@)T`
`10 sqrt(3) = 10 xx (sqrt(3))/(2)T rArr T = 2sec`.
(b) `v_(x) = u_(x) +a_(x) t rArr v_(x) = g cos 60^(@) xx T`
`v_(x) = 5 xx 2 rArr v_(x) = 10 m//s`
`v = 10 m//s`

(c ) `x = x_(0) +u_(x)t +(1)/(2) a_(x)t^(2) rArr OP = (1)/(2) (g cos 60^(@))2^(2)`
`OP = (1)/(2) xx ((g)/(2))2^(2) rArr OP = 10m`
`(h)/(OP) = sin 30^(@) rArr h = 5m`
(d) `H = h +((u sin 60^(@))^(2))/(2g) rArr H - 5 +((10sqrt(3)+(sqrt(3))/(2)))/(20)^(2)`
`H = 5+ ((15)^(2))/(20) rArr H = 5+(225)/(20)`
`H = 5+(45)/(4) rArr H = 5 xx 11.25`
`H = 16.25 m`
(e) for distance `PQ`
`y = u_(y)t +(1)/(2) a_(y) t^(2)`
`rArr OQ = 10 sqrt(3) xx2 +(1)/(2) (-g sin 60^(@)) xx 2^(2)`
`OQ = 10 sqrt(3) xx 2+(1)/(2) xx (-(10sqrt(3))/(2)) xx 2^(2)`
`OQ = 20 sqrt(3) - 10 sqrt(3)`
`OQ = 10 sqrt(3) rArr PQ = sqrt(OP^(2) +OQ^(2))`
`PQ = sqrt(10^(2) +(10sqrt(3))^(2)) rArr PQ = sqrt(400)`
`PQ = 20 cm`