Answer:
Speed of the plane over the ground is 429.33 km/hr due West and 70 km/hr towards North
Solution:
According to the question:
Wind flowing in South direction, [tex]v_{s} = 70.0 km/h[/tex]
Air speed, [tex]v_{a} = 435.0 km/h[/tex]
Now,
The velocity vector is required 70 km/h towards north in order to cancel the wind speed towards south.
Therefore,
The ground speed of the plane is given w.r.t fig 1:
[tex]v_{pg} = \sqrt{v_{a^{2}} - v_{s}^{2}}[/tex]
[tex]v_{pg} = \sqrt{435.0^{2} - 70.0^{2}} = 429.33 km/h[/tex]
