A particle A is projected with an initial velocity of 60 m/s at an angle `30^(@)` to the horizontal. At the same time a second particle B is projected in opposite direction with intial speed of 50 m/s from a point at a distance of 100 m from A. if the particles collide in air, find a the angle of projection `alpha` of particle B (b) time when the collisions takes place and c the distance of P from A. where collision occurs `(g=10m//s^(2))`

Explanation

a. Taking x and y directions as shown in figure
Here `veca_(A)=-ghatj`
`veca_(B)=-ghatj`
`u_(Ax)=60cos 30^(@)=30sqrt(3)` m/s `u_(Ay)=60 sin 30^(@)=30m//s`
`u_(Bx)=-50cos alpha` and `u_(By)=50sin alpha`

Relative acceleration between the two is zero as `veca_(A)=veca_(B)`. Hence the relative motion between the two is uniform. Condition of collision is that `vecu_(AB)` should be along AB this is possible only when `u_(Ay)=u_(By)` i.e. component of relative velocity along y-axis should be zero.
(or) `30=30sin alpha :. =sin^(-1)(3//5)`
b. Now `|vecu_(AB)|=u_(Ax)-u_(Bx)`
`=(30sqrt(3)+50cos alpha)m//s=(30sqrt(3)+50xx4/3)m//s=(30sqrt(3)+40)m//s`
Therefore time of collision is `t=(AB)/(|vecu_(AB)|)=100/(30sqrt(3)+40)=1.09s`
c. Distance of point P from A where collisioni takes place is
`s=sqrt((u_(Ax)t)^(2)+(u_(Ay)t-1/2"gt"^(2))^(2))=sqrt((30sqrt(3)xx1.09)^(2)+(30xx1.09-1/2xx10xx1.09xx1.09)^(2)):.s=62.64m`