Question

A stone is projected from a point on the ground in such a direction as to hit a bird on the top of a telegraph post of height $h$ and then attain the maximum height $3h/2$ above the ground. If at the instant of projection, the bird were to fly away horizontally with uniform speed, find the ratio between the horizontal velocities of the bird and the stone if the stone still hits the bird while descending. Take $\sqrt{3}=1.73$.

Answer 1

From motion from A to B, time of flight
$t_2-t_1=\frac{2v_y}{g}---\left(1\right)$
where $v_y$ is the vertical velocity at $t_1$
Using second equation of motion,
$h=u_{y}t-\frac{1}{2}g{t}^2\Rightarrow g{t}^2-2u_{y}t+2h=0$
The above equation has the roots $t=t_1$ and $t=t_2$.
$\therefore t_1{t}_2=\frac{2h}{g}----\left(2\right)$
Also,maximum height attained from point A,
$\frac{h}{2}=\frac{v_y^2}{2g}\Rightarrow v_y=\sqrt{gh}----\left(3\right)$
$\Rightarrow t_2-t_1=2\sqrt{\frac{h}{g}}\text{(from (1))}----\left(4\right)$
We know $(t_2-t_1{)}^2=(t_1+t_2{)}^2-4t_1{t}_2$
$\Rightarrow \frac{4h}{g}=(t_1+t_2{)}^2-\frac{8h}{g}$
$\Rightarrow t_1+t_2=2\sqrt{\frac{3h}{g}}----\left(5\right)$
From (4 ) & (5), $t_2=\sqrt{\frac{h}{g}}(\sqrt{3}+1)$
If the velocity of the bird is $V_b$,
$AB=V_b{t}_2=u_x(t_2-t_1)$
$\therefore \frac{V_b}{u_x}=\frac{t_2-t_1}{t_2}=\frac{2}{\sqrt{3}+1}=\frac{2(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}$
$=\sqrt{3}-1$
$=1.73-1=0.73$