A ball is thrown from a point on a 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 ${}^{\prime }{u}^{\prime }$. Given that during its flight, the ball just touches the bird at only one point. Find the distance on the ground from the point of projection where the ball strikes.

The correct option is A $2u\sqrt{\frac{2h}{g}}$
Given that,
vertical distance of bird from ground $=h$
speed of bird (horizontal) $=u$
The ball will touch the bird only if horizontal velocity of ball = horizontal velocity of the bird.
$\Rightarrow u_x=u....\left(1\right)$
Now, vertical component of the velocity of ball is $u_y$
If the ball strikes the bird at exactly one point, it has to strike at the maximum height of its trajectory.
i.e $h=\frac{u_y^2}{2g}$
$\Rightarrow u_y=\sqrt{2gh}$
Now, range of ball is
$R=\frac{2u_x{u}_y}{g}$
$\Rightarrow R=\frac{2u_x\sqrt{2gh}}{g}$
$\Rightarrow R=2u\sqrt{\frac{2h}{g}}$.
Hence, option (a) is correct.