Question

The airspeed of an airplane is 600 km/hr, and there is a wind blowing northeast at 90 km/hr. If the airplane needs to head due north, at what angle does the pilot need to fly the plane?

Answer 1

Answer:

The pilot has to fly the plane at an angle 83° north of west.

Explanation:

The airspeed of an airplane is 600 km/hr ( north direction)

In figure, $V_{PA}=600\text{ km/h}$

A wind blowing northeast at 90 km/h

In figure,  $V_{A}=90\text{ km/h}$

Let actual speed of airplane and direction of plane,

Speed = x km/h

Direction = Ф (north of west, refer figure)

Now, we divide each velocity in horizontal and vertical component.

• For wind, $V_{A}=90\text{ km/h}$

Horizontal component, $V_{\text{Horizontal}}=90\cos45^\circ\text{ km/h}$

Vertical component, $V_{\text{Vertical}}=90\sin45^\circ\text{ km/h}$

• For airplane,

Horizontal component, $V_{\text{Horizontal}}=x\cos\theta\text{ km/h}$

Vertical component, $V_{\text{Vertical}}=x\sin\theta\text{ km/h}$

If the airplane needs to head due north.

Horizontal component must be cancel out.

Therefore, $x\cos\theta=90\cos45^\circ----------(1)$

Speed of airplane is 600 km/h due to north.

So, Sum of vertical component must be equal to 600

Therefore, $x\sin\theta+90\sin45^\circ=600-----------(2)$

Using eq(1) and eq(2) we get,

$\tan\theta=\dfrac{600\sqrt{2}-90}{90}$

$\theta=83.23^\circ$

$\theta\approx 83^\circ$

Hence, The pilot has to fly the plane at an angle 83° north of west.