# An airplane pilot sets a compass course due west and maintains an air speed of ( 240 mathrm{km} / mathrm{h} ). After flying for ( frac{1}{2} h, ) he finds himself over a town that is ( 150 mathrm{km} ) west and ( 40 mathrm{km} ) south of his starting point. The wind velocity (with respect to ground) is:

(A) ( 100 mathrm{km} / mathrm{h}, 37^{circ} mathrm{W} ) of ( mathrm{S} )

(B) ( 100 mathrm{km} / mathrm{h}, 37^{circ} mathrm{S} ) of ( mathrm{W} )

(C) ( 120 mathrm{km} / mathrm{h}, 37^{circ} mathrm{W} ) of ( mathrm{S} )

(D) ( 120 mathrm{km} / mathrm{h}, 37^{circ} mathrm{S} ) of ( mathrm{W} )

As the pilot is ( 150 mathrm{km} ) west and ( 40 mathrm{km} ) south. So, we can obtain the position of pilot using pythagorus theorem.

( d^{2}=(150)^{2}+(40)^{2} )

( =22500+1600 )

( =24100 mathrm{km}^{2} )

( d=sqrt{24100}=155 mathrm{km} )

( theta=tan ^{-1}left(frac{40}{150}right)=tan ^{-1} 0.2667 )

( =15^{circ} )

Velocity of piolet with respect to ground ( left(vec{v}_{1}right) ) is given by ( vec{v}_{1}=frac{155}{0.5}=310 mathrm{km} mathrm{h} )

Let ( overrightarrow{mathrm{v}}_{2} ) be the velocity of air and Let ( vec{v}_{3} ) be the velocity of wind with respect to ground. Now, Using cosine law we have ( overrightarrow{mathrm{v}}_{3}^{2}=vec{v}_{1}^{2}+overrightarrow{mathrm{v}}_{2}^{2}-2 vec{v}_{1} vec{v}_{2} cos 15^{circ} )

( =(240)^{2}+(310)^{2}-2(210)(310)(0.9659) )

( =57600+96100-1437 )

( =9970 )

( Rightarrow overrightarrow{mathrm{v}}_{3}=100 mathrm{km} mathrm{h} )