Question

A stone is projected from the point of a ground in such a direction so as to hit a bird on the top of a telegraph post of height h and then attain the maximum height 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 velocities of the bird and the stone, if the stone still hits the bird while descending:

• A. $\frac{1}{[\sqrt{2}-1]}$
• B. $\frac{[\sqrt{2}-1]}{2}$
• C. $2[\sqrt{2}-1]$
• D. $\frac{2}{[\sqrt{2}+1]}$

Answer 1

The correct option is D $\frac{2}{[\sqrt{2}+1]}$

Let $u$ is the initial velocity

Maximum height $H=\frac{u^2{\sin }^2\theta }{2g}=2h$

$h=\frac{u^2{\sin }^2\theta }{4g}$

Horizontal component of the velocity $u_x=u\cos \theta$

Vertical component $u_y=u\sin \theta$

At maximum height

$v_y^2-u_y^2=-2g\left(2h\right)$

$u_y=u\sin \theta =2\sqrt{gh}$

$t$ be the time taken by the stone to attain the vertical height $h$

$h=u_{y}t-\frac{1}{2}g{t}^2$

$t^2-\left(\frac{2u_y}{g}\right)t+\frac{2h}{g}=0$

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

$t_1=(4-\sqrt{2})\sqrt{\frac{h}{g}}$

$t_2=(4+\sqrt{2})\sqrt{\frac{h}{g}}$

So bird travel horizontal distance from initial position to final position = AB in time $t_2$

velocity of the bird$=v_b$

$AB=v_b{t}_2$

Same distance AB traveled by the stone in time $t_2-t_1$

$AB=u_x(t_2-t_1)$

$\frac{v_b}{u_x}=\frac{(t_2-t_1)}{t_2}=\frac{\left(\right(4+\sqrt{2}\left)\sqrt{\frac{h}{g}}\right)-\left(\right(4-\sqrt{2}\left)\sqrt{\frac{h}{g}}\right)}{(4+\sqrt{2})\sqrt{\frac{h}{g}})}=\frac{2}{(\sqrt{2}+1)}$