Question

$A,B,C$ and $D$ are four trees, located at the vertices of a square. Wind blows from $A$ to $B$ with uniform speed. The ratio of times of flight of a bird from $A$ to $B$ and from $B$ to $A$ is n. At what angle should the bird fly from the direction of wind flow, in order that it starts from $A$ and (a) reaches $C$, (b) reaches $D$?

Answer 1

Let the velocity of bird w.r.t. air is v and the velocity of wind is u.

The velocity of bird from A to B,

$|\overrightarrow{v_{AB}}|=v+u$

Velocity of bird from B to A,

$|\overrightarrow{v_{BA}}|=v-u$

$t_{AB}=\frac{l}{v+u},t_{BA}=\frac{l}{v-u}$

$\frac{t_{AB}}{t_{BA}}=n\Rightarrow n=\frac{v-u}{v+u}$

$\frac{u}{v}=\frac{1-n}{1+n}$

a. Motion of bird from A to C

$\overrightarrow{v_{bird}}=\overrightarrow{v_{bird.wind}}+\overrightarrow{v_{wind}}=\overrightarrow{v}+\overrightarrow{u}$

From velocity diagram, we can use sine rule:

$\frac{u}{sin\theta }=\frac{v}{sin{45}^o}$

$\Rightarrow sin\theta =\frac{u}{\sqrt{2}v}=\frac{1-n}{\sqrt{2}(1+n)}$

$sin(\alpha -sin{45}^o)=\frac{1-n}{\sqrt{2}(1+n)}$

$sin\alpha cos{45}^o-cos\alpha sin{45}^o=\frac{1-n}{\sqrt{2}(1+n)}$

$sin\alpha -cos\alpha =\frac{1-n}{1+n}$

Squaring both sides,

$1-2sin\alpha cos\alpha ={\left(\frac{1-n}{1+n}\right)}^2$

$si{n}^2\alpha =\frac{4n}{(1+n{)}^2}$

$\Rightarrow =\frac{1}{2}si{n}^{-1}\left[\frac{4n}{(1+n{)}^2}\right]$

For motion of bird from A to D: From vector diagram, it is clear

$sin{\theta }^{\prime }=\frac{u}{v}=\frac{1-n}{1+n}$

$sin(\alpha -{90}^o)=\frac{1-n}{1+n}$

$cos\alpha =-\frac{(1-n)}{1+n}$

$\alpha =co{s}^{-1}\left[\frac{-(1-n)}{1+n}\right]$