Question

An airplane pilot wishes to fly due west. A wind of 70.0 km/h is blowing toward the south. And I need to find out the speed of the plain over the ground. It is given that the speed is in still air and the airspeed is 435.0 km/h.

Answer 1

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,
`v_(s) = 70.0 km/h`

Air speed,
`v_(a) = 435.0 km/h`

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:

`v_(pg) = \sqrt{v_(a^(2)) - v_(s)^(2)}`

`v_(pg) = \sqrt{435.0^(2) - 70.0^(2)} = 429.33 km/h`