Question

In a harbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the north east direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?

Answer 1

see figure, here it is clearly shown that velocity of wind and the boat.

a/c to question,
Velocity of wind , $v_w$ = 72 km/hr along N-E direction
or, $\vec{v_w}=72cos45\hat{i}+72sin45\hat{j}$

Velocity of boat , $v_b$ = 51 km/hr along North direction
$\vec{v_b}=51\hat{j}$

To find out the direction of flag on the mast of the boat, we need to find out the relative velocity of wind w.r.t the boat.

Relative velocity of wind w.r.t the boat , $v_{wb}=v_w-v_b$

= $72cos45\hat{i}+72sin45\hat{j}-51\hat{j}$
=$36\sqrt{2}\hat{i}+(36\sqrt{2}-51)\hat{j}$

actually, angle between velocity of wind and velocity of boat is 135° [ see figure ]

Using parallelogram law of vectors, let $v_{wb}$ be the resultant vector making an angle β with $v_w$

so, $tan\beta=\frac{V_bsin135^{\circ}}{V_w+V_bcos45^{\circ}}$

= 51 × sin135°/(72 + 51cos45°)

= 1.0034

β = 45.099° ≈ 45.1° wrt N-E direction .
hence, the direction of flag on the mast of the boat is 0.1˚ w.r.t East.

Answer 2

Your answer is here with the attachement