A helicopter files 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 `n``u` where `n gt 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.
.

Explanation

Velocity of helicopter w.r.t. groundn is given by
`vec V_(Helicopter, air) = vec V_(Helicopter,air) + vec V _(Air) = vec V_(Air) + vec V_(Helicopter, air)`
`|vec V_(H A)| = n u, |vec V_A| = u, |vec 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 - 2 u v cos theta`
or `v^2 - 2 u v 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 - sin^2 theta)`
`v - u cos theta = +- u sqrt(n^2 - sin^2 theta)`
`v = u cos theta +- 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 helicoper 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 consine rule inthe vector triangle, we get
`n^2 u^2 = u^2 + V^2 - 2u V cos (180^@ - theta)`
=`u^2 + V^2 + 2u V cos theta`
`V = -u cos theta + u sqrt(n^2 - sin^2 theta)`
The total time for motion `A` to `B` and `B` to `A`,
`t = (d)/(v) + (d)/(V)`
=`(d (u + V))/( uV)`
=`(2 d u sqrt(n^2 - sin^2 theta))/(u^2 (n^2 - sin^2 theta) - u^2 cos^2 theta)`
=`(2 d u sqrt(n^2 - sin^2 theta))/(u^2 (n^2 - 1)) = (2 d sqrt(n^2 - sin^2 theta))/(u(n^2 - 1))`.
, .