Two inclined planes OA and OB having inclination (with horizontal) `30^(@)` and `60^(@)`, respectively, intersect each other at O as shown in figure. A particle is projected from point P with velocity `u = 10(sqrt3) ms^(-1)` along a direction perpendicular to plane OA. If the particle strikes plane OB perpendicularly at Q, calculate

The velocity with which particle strikes the plane OB,

Explanation

In the shown co-ordnate system for motion of the particle along x-axis .
` U_(x) = u`
` a_(x) = -g sin 60^(@)`
For motion of the paricle along y-axis.
` U_(y) = 0, a_(y) = - g cos 60^(@)`
(i) If particle strikes the plane ON after time t then ` v_(x) = u_(x) + a_(x) t`
Now ` v_(x) =0 ` therefore ,
` Rightarrow t = u/(g sin 60^(@))`
` = ( 10sqrt3)/(1 0xx sqrt3/2)`
= 2s
(ii) For velocity of the particle after time t = ( 2s)
` v_(y) = u_(y) +a_(y)t`
= ` 0 - gcos 60^(@) xx 2`
` - 10xx 1/2 xx2`
10 m/s
(iii) For distance AO
` y_(AO) |u_(y) + 1/2a_(y)t^(2)|`
`|0+ 1/2 ( - g cos 60^(@)t^(2)|`
` |-1/2 xx 10 xx 1/2 xx 2^(2)|`
= | - 10 m|
10 m
Now height A from the horizontal level of O
`h = OA sin 30^(@)`
` 10 xx 1/2`
(iv) For distance OB
Considering motion along x-axis
` v_(x)^(2) = u_(x)^(2) + 2a_(x) (OB) Rightarrow 0 = ( 10sqrt3)^(2) - 2g sin 60^(@) xx ( OB)`
`Rightarrow 0 =300 - 2 xx 10 xx sqrt3/2 xx ( OB)`
` Rightarrow OB = 10sqrt3 ` m
Now , from right angle ` triangle AOB`
Distance AB = ` sqrt(OA^(2) +OB^(2))`
` = sqrt(10^(2) + (10sqrt3)^(2))`
= 20m