Two particles A and B are projected simultaneously in the directins shown in figure with velocities `v_(A) = 20 m//s and v_(B) = 10 m//s` respectively. They collide in air after `1/2` s. Find
(a) the angle `theta`
(b) the distance x.
.

Explanation

Both particle will collide if they are at same height in same time.
`y_(A)=y_(B)rArr (u_(y))_(A)t-(1)/(2)"gt"^(2)=(u_(y))_(B)t-(1)/(2)"gt"^(2)rArr(u_(y))_(A)=(u_(y))_(B)`
`rArr(V_(A)sin theta)=v_(B)rArr 20 sin theta=10rArr sintheta=(1)/(2)rArrtheta=30^(@)`
In 0.5s horizontal distance covered by A is `x=(u_(x))_(A)t=(20 cos 30^(@))0.5=10xxsqrt(3)/(2)=5sqrt(3)m`