Question

A helicopter flies horizontally with constant velocity in a direction $\theta$ east of north between two points $A$ and $B$, at distance d apart. Wind is blowing from south with constant speed u; the speed of helicopter relative to air is nu, where n > 1. Find the speed of the helicopter along AB. The helicopter returns from B to A with same speed nu relative to air in same wind. Find the total time for the journeys.

Answer 1

Velocity of helicopter w.r.t. ground is given by

${\overrightarrow{V}}_{Helicopter}={\overrightarrow{V}}_{Helicopter.air}+{\overrightarrow{V}}_{air}={\overrightarrow{V}}_{air}+{\overrightarrow{V}}_{Helicopter.air}$

$|{\overrightarrow{V}}_{HA}|=nu,|{\overrightarrow{V}}_A|=u,|{\overrightarrow{V}}_H|=V$

Hence, the actual velocity of helicopter is the vector sum of its velocity of helicopter relative to air and velocity of air.

Using cosine rule, for motion from A to B,

$n^2{u}^2=u^2+v^2-2uv\cos \theta$

or $v^2-2uv\cos \theta =u^2(n^2-1)$

or $(v-u\cos \theta {)}^2=u^2(n^2-1)+u^2{\cos }^2\theta$

$=u^2(n^2-si{n}^2\theta )$

$v-u\cos \theta =\pm u\sqrt{n^2-{\sin }^2\theta }$

$v=u\cos \theta \pm u\sqrt{n^2-{\sin }^2\theta }$

As $n\ge 1$,

$n^2-{\sin }^2\theta \ge 1-{\sin }^2\theta \ge {\cos }^2\theta$

Hence, speed of helicopter is $v=u\cos \theta +u\sqrt{n^2-{\sin }^2\theta }$

Now we consider motion from B to A.

Similar to first case, the helicopter will return to A along BA when its velocity after being deflected by wind is along BA.

Once again using cosine rule in the vector Na triangle, we get

$n^2{u}^2=u^2+V^2-2uVcos({180}^o-\theta )$

$=u^2+V^2+2uV\cos \theta$

$V=-ucos\theta +u\sqrt{n^2-si{n}^2\theta }$

The total time for motion A to B and B to A,

$t=\frac{d}{v}+\frac{d}{V}$

$=\frac{d(u+V)}{uV}$

$=\frac{2du\sqrt{n^2-si{n}^2\theta }}{u^2(n^2-si{n}^2\theta )-u^{2}co{s}^2\theta }$

$=\frac{2du\sqrt{n^2-si{n}^2\theta }}{u^2(n^2-1)}=\frac{2d\sqrt{n^2-si{n}^2\theta }}{u(n^2-1)}$