A ball is thrown from a point on ground at some angle of projection. At the same time a bird starts from a point directly above this point of projection at a height h horizontally with speed u. GHiven that in its flight ball just touches bird at one point. Find the distance on ground where ball strikes.

Dear Vishrant

As the question says the ball touches the bird at one point.So that point must be the highest point attained by the ball.Because if it was not the highest point then the ball must touch the bird once again i.e when the ball is moving down.

let v be the velocity of ball at the time of projection and θ be the angle of projection

then since horizontal component of velocity of ball = velocity of bird (since they both cover same horizontal distance in same time)

v cos θ =u -eq 1

futher h= v² sin²  θ/2g

√h= v sinθ /√2g -eq 2

distance on ground where ball strikes=rnage of the projectile=v² sin 2θ/g=2v² sin θ cos θ /g

multiply eq 1 and eq 2

u√h=v²sin θ cosθ/√2g

2u√2h/√g=v² sinθ cosθ*2/g=range

so ball will land on ground at a distance of 2u√2h/√g