Question

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.

• A. $2u\sqrt{\frac{2h}{g}}$
• B. $2u\sqrt{\frac{h}{g}}$
• C. $u\sqrt{\frac{h}{g}}$
• D. $u\sqrt{\frac{h}{2g}}$

Answer 1

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.