A-stone-is-projected-from-a-point-on-the-ground-in-such-a-direction-so-as-to-hit-a-bird-on-the-top-of-a-telegraph-post-of-height-h-and-then-attain-a-height-2h-above-the-ground-If-at-an-instant-of-p

Question Number 18131 by Tinkutara last updated on 15/Jul/17

A stone is projected from a point on the

ground in such a direction so as to hit a

bird on the top of a telegraph post of

height h, and then attain a height 2 h

above the ground . If, at an instant of

projection, the bird were to fly away

horizontal with a uniform speed, find

the ratio of the horizontal velocities of

the bird and the stone, if the stone still

hits the bird .

Commented by Tinkutara last updated on 15/Jul/17

Commented by ajfour last updated on 15/Jul/17

y = xtan θ - \frac{{gx}^{2}}{{2 u}^{2} \cos^{2} θ}

let m = \tan θ, θ the ∠ of projection .

x = d, and and x = R - d when y = h

d is the distance to telegraph post

from point of projection . R is the

range of stone . So

h = mx - \frac{{gx}^{2}}{2} (1 + m^{2})

g (1 + m^{2}) x^{2} - 2 mx + 2 h = 0

roots are x = d and x = R - d

\Rightarrow sum of roots = R = \frac{2 m}{g (1 + m^{2})}

product = d (R - d) = \frac{2 h}{g (1 + m^{2})}

\Rightarrow \frac{d (R - d)}{R} = \frac{h}{m} \dots . (i)

2 h = \frac{u^{2} \sin^{2} θ}{2 g} and R = \frac{{2 u}^{2} \sin θ \cos θ}{g}

dividing equations we get

\frac{8 h}{R} = m \Rightarrow \frac{h}{m} = \frac{R}{8}

substituting for \frac{h}{m} in (i)

\frac{d (R - d)}{R} = \frac{R}{8}

\Rightarrow \frac{d}{R} (1 - \frac{d}{R}) = \frac{1}{8}

{(\frac{d}{R})}^{2} - (\frac{d}{R}) + \frac{1}{8} = 0

\Rightarrow {(\frac{d}{R} - \frac{1}{2})}^{2} = \frac{1}{8} \Rightarrow R - 2 d = \frac{R}{\sqrt{2}}

R (1 - \frac{1}{\sqrt{2}}) = 2 d

let v be the bird velocity

\frac{v}{ucos θ} = \frac{R - 2 d}{R - d} = \frac{1 - \frac{2 d}{R}}{1 - d / R}

= \frac{1 - (1 - \frac{1}{\sqrt{2}})}{1 - (\frac{1}{2} - \frac{1}{2 \sqrt{2}})} = \frac{(\frac{1}{\sqrt{2}})}{\frac{1}{2} (1 + \frac{1}{\sqrt{2}})}

= \frac{2}{\sqrt{2} + 1} .

Commented by Tinkutara last updated on 15/Jul/17

Thanks Sir!