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 initial 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 collision takes place and (c) the distance of P from A, where collision occurs. `(g= 10 m//s^2)`

Explanation

Taking horizontal as x-axis, vertical as y-axis and A as origin
` vecu_(A) = ( 60 cos 30^(@)) hati + (60sin 30^(@)hatj`
` = 30 sqrt3 hati + 30 hatj`
`vecu_(B) = ( - 50 cos alpha)hati ( 50 sin alpha)hatj`
As both the particles are falling under gravity. their relative acceleration is zero.
Now initial velocity of second particle w.r.t first particle
` vecc_("BA") = vecv_(B) - vecv_(A)`
` ( - 30 sqrt3 - 50 cos alpha) hati + ( 50 sin alpha -30) hatj`
As acceleration of B w.r.t A is zero so B will appear to move in straight line with uniform speed. FOr collision to take palce, velocity , of B ,w,r,t A must be toward , A, i.e. y component of relative velocity must be zero.
` ( 50 sin alpha -30) =0`
` sin alpha = 3/5 Rightarrow alpha = sin^(-1) (3/5)`
(ii) Time of colision
` vecd_(AB) = vecV_(AB) (T)`
`-100 hati = ( -30 sqrt3 - 50 cos alpha) I hati`
`or 100 = ( 30 sqrt3 + 50 xx 4/5) T`
` or T = 100/(( 30sqrt3 + 40)) = 10/( 3 sqrt3 + 4) = 1.09 s `