A light plane attains an airspeed of 500 km/h. The pilot sets out for

To solve this problem, we can break down the velocity of the plane into two components: the northward component and the eastward component. Let's denote the magnitude of the northward component as $v_n$ and the magnitude of the eastward component as $v_e$. We also need to find the magnitude and direction of the wind velocity, denoted as $v_w$ and ${\theta }_w$ respectively.
The northward component of the plane's velocity can be calculated using the equation:
$v_n=v_p\cos ({\theta }_p)$
where $v_p$ is the airspeed of the plane (500 km/h) and ${\theta }_p$ is the angle the plane is heading with respect to due north (20.0 degrees east of due north). Converting the angle to radians, we have ${\theta }_p=\frac{20\pi }{180}$.
Substituting the given values, we get:
$v_n=500\cos \left(\frac{20\pi }{180}\right)$
The eastward component of the plane's velocity can be calculated using the equation:
$v_e=v_p\sin ({\theta }_p)$
Substituting the values, we have:
$v_e=500\sin \left(\frac{20\pi }{180}\right)$
Since the wind is affecting the plane's velocity, we can express the overall velocity of the plane in terms of the wind velocity:
$v_{\text{total}}=v_p+v_w$
where $v_{\text{total}}$ is the magnitude of the total velocity of the plane and $v_w$ is the magnitude of the wind velocity.
Given that the plane traveled a distance of 800 km due north in 2.00 hours, we can calculate the magnitude of the total velocity of the plane:
$v_{\text{total}}=\frac{800}{2.00}$
Now, we can equate the components of the total velocity and the individual components of the plane's velocity and wind velocity:
$v_n=\frac{800}{2.00}$
$v_e=0$
$v_w=\frac{800}{2.00}-v_p$
Finally, we can calculate the direction of the wind velocity using the tangent function:
$\tan ({\theta }_w)=\frac{v_e}{v_n}$
Now let's substitute the values and calculate the solution:
The magnitude of the northward component of the plane's velocity:
$v_n=500\cos \left(\frac{20\pi }{180}\right)\approx 472.92\text{km/h}$
The magnitude of the eastward component of the plane's velocity:
$v_e=500\sin \left(\frac{20\pi }{180}\right)\approx 170.92\text{km/h}$
The magnitude of the total velocity of the plane:
$v_{\text{total}}=\frac{800}{2.00}=400\text{km/h}$
The magnitude of the wind velocity:
$v_w=\frac{800}{2.00}-500=-100\text{km/h}$
The direction of the wind velocity:
$\tan ({\theta }_w)=\frac{v_e}{v_n}$
${\theta }_w=\arctan \left(\frac{v_e}{v_n}\right)$
Substituting the values:
${\theta }_w=\arctan \left(\frac{170.92}{472.92}\right)\approx {20.42}^{\circ }$
Therefore, the (a) magnitude of the wind velocity is $100$ km/h (opposite direction of the plane's motion) and (b) the direction of the wind velocity is ${20.42}^{\circ }$ east of due south.