Question

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

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

Answer 1

The correct option is C $2u\sqrt{\frac{2h}{g}}$

As at one point the ball touches the bird it simply means that $\text{Horizontal velocity of the ball which is constant over time is EQUAL with that of the bird and as the ball}$

$\text{JUST TOUCHES one time it simply menas that it will be at its top point during that moment so that}$

$\text{it never become at that height again}$

So if $vcos\theta$ is the horizontal velocity of the ball then $u=vcos\theta$ and $h=\frac{(vSin\theta {)}^2}{2g}$ or $vsin\theta =\sqrt[2]{22gh}$

Where $v$ is projection velocity and $\theta$ is projection angle.

Now the question ask the horizontal range $R=\frac{2vsin\theta }{g}\times vcos\theta =\frac{2\times \sqrt[2]{22gh}}{g}\times u=2u\sqrt[2]{2\frac{2h}{g}}$

Option C is correct.