Question

A helicopter is to reach a point 200,000 m east of his existing place. Its velocity relative to wind blowing at 30 kmh^-1 from north west taking scheduled arrival time duration as 40 minute is

• A. $21\widehat{i} + 279\widehat{j}$
• B. $279\widehat{i} + 21\widehat{j}$
• C. $729\widehat{i} + 12\widehat{j}$
• D. $12\widehat{i} + 729\widehat{j}$

Answer 1

Answer: (B)

$279\widehat{i} + 21\widehat{j}$

Answer 2

Velocity of wind w.r.t. earth is,

v_WE = 30 kmh^-1 towards NS i.e., SE velocity of helicopter w.r.t. earth is

v_HE = $\frac{200000}{1000 \times (40/60)}$

= 300 km towards East

In vector form,

$\overrightarrow{v}$_HE = 300$\widehat{i}$ and

$\overrightarrow{v}$_WE = (30 cos 45⁰)$\widehat{i}$ - (30 sin 45⁰)$\widehat{j}$

= 15 $\sqrt { 2 }$ $\widehat{i}$ - 15 $\sqrt { 2 }$ $\widehat{j}$

Then resultant velocity is also given by, $\overrightarrow{v}$_NH = $\overrightarrow{v}$_WE + $\overrightarrow{v}$_HW

or

$\overrightarrow{v}$_HW = $\overrightarrow{v}$_HE - $\overrightarrow{v}$_WE

= 300$\widehat{\mathbf{i}}$- (15√2$\widehat{\mathbf{i}}$ - 15√2$\widehat{j}$)

= 279$\widehat{i}$ + 21$\widehat{j}$