From points `A` and `B`, at the respective heights of `2m` and `6m`, two bodies are thrown simultaneously towards each other, one is thrown horizontally with a velocity of `8m//s` and the other, downward at an angle `45^(@)` to the horizontal at an initial velocity `v_(0)` such that the bodies collide in flight. This horizontal distance between points `A` and `B` equal to `8m`.Then find
a)The initial velocity `V_(0)` of the body thrown at an angle `45^(@)`
b)The time of flight `t` of the bodies before colliding
c) The coordinate `(x,y)` of the point of collision (consider the bottom of the tower `A` as origin )is

Explanation

a) From diagram
`tan theta=4/8 rArr tan theta=1/2`…..(1)
`vecv_(A)=8hati,v_(B)= - vec(v)_(0)cos45^(@)hati-vecv_(0)sin45^(0)hatj`
`vecv_(BA)=(-v_0/sqrt2-8)i-v_(0)/sqrt2j`
Direction of `vecv_(BA)`
`tan theta=v_(0)/sqrt2sqrt2/((v_(0)+8sqrt2)`....(2)
From eq (1) and eq (2) `2v_(0)=v_(0)+8sqrt2,v_(0)=11.28m//s`
`vecv_(BA)=(-v_0/sqrt2-8)i-v_(0)/sqrt2j`
`:' v_(0)=8sqrt2Ϸvec v_(BA)=-16hati-8hatj`
`|vec v_(BA)|=t=S_(BA)`
`(sqrt((-16)^(2)+(-8)^(2)))t=sqrt(8^(2)+4^(2))`
`t=sqrt(80/320)=sqrt(1/4) rArr t=0.5s`
c)`x=v_(x)t=(8)(0.5)=4`
`y'=1/2"gt"^(2)=1/2xx10xx1/4=1.25`
`y=2-y'=0.75`