Question

A stone is projected from the ground in such a direction so as to hit a bird on the top of a telegraph post of height $h$ and attains the maximum height of $2h$ above the ground. If at the instant of projection, the bird were to fly away horizontally with a uniform speed, find the ratio between the horizontal velocity of bird and the horizontal component of velocity of stone, if the stone hits the bird while descending.

• A. $\frac{2}{\sqrt{2}+1}$
• B. $\frac{3}{\sqrt{2}+1}$
• C. $\frac{2}{\sqrt{3}+1}$
• D. $\frac{3}{\sqrt{3}+1}$

Answer 1

The correct option is A $\frac{2}{\sqrt{2}+1}$

Let the initial velocity of the stone be $u$ at an angle of $\theta$ with the horizontal.

Since the stone reaches a maximum height of $2h$,

$0^2-u^{2}si{n}^2\theta =2(-g)\left(2h\right)$

$⟹\sin \theta =\frac{2\sqrt{gh}}{u}$

Let the stone hit the bird flying with velocity $v$ at time $t$.

Thus $h=u\sin \theta t-\frac{1}{2}g{t}^2$

$⟹h=2\sqrt{gh}t-\frac{1}{2}g{t}^2$

$⟹t=(4\pm \sqrt{2})\sqrt{\frac{h}{g}}$

The distance traveled by bird from initial position to final position=AB in time $t_2=(4+\sqrt{2})\sqrt{\frac{h}{g}}$

$AB=u\cos \theta t_2=v(t_2-t_1)$

Thus $\frac{u\cos \theta }{v}=\frac{t_2-t_1}{t_2}=\frac{2}{\sqrt{2}+1}$