A plane is flying at $$ 300 \mathrm{~m} / \mathrm{s} $$ in the $$ y $$-direction. The wind is blowing at $$ 20.0 \mathrm{~m} / \mathrm{s} $$ in the $$ x $$-direction. What is the direction of the velocity of the plane? $$ \theta=[?]^{\circ} $$

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1. The plane is flying with a velocity of $$300\,\mathrm{m/s}$$ in the $$y$$-direction. Its velocity vector is  
   $$
   \vec{v}_\text{plane} = (0,\,300).
   $$

2. The wind is blowing with a velocity of $$20.0\,\mathrm{m/s}$$ in the $$x$$-direction. Its velocity vector is  
   $$
   \vec{v}_\text{wind} = (20.0,\,0).
   $$

3. The resultant (ground) velocity of the plane is the vector sum of its own velocity and the wind velocity:  
   $$
   \vec{v}_\text{resultant} = \vec{v}_\text{plane} + \vec{v}_\text{wind} = (20.0,\,300).
   $$

4. To find the direction of the resultant velocity, we compute the angle $$\theta$$ that the velocity vector makes with the $$y$$-axis. Here, the $$x$$-component comes from the wind and the $$y$$-component from the plane’s own motion. The angle can be found by  
   $$
   \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{20.0}{300}.
   $$

5. Calculate the angle:  
   $$
   \theta = \tan^{-1}\left(\frac{20.0}{300}\right) = \tan^{-1}(0.06667).
   $$

6. Evaluating the inverse tangent, we find:  
   $$
   \theta \approx 3.81^\circ.
   $$

Thus, the direction of the plane's velocity is approximately  
$$
\theta \approx 3.81^\circ
$$
from the $$y$$-axis (i.e., $$3.81^\circ$$ east of north).
The plane's velocity direction is approximately $$3.81^\circ$$ east of north.

A plane is flying at \( 300 \mathrm{~m} / \mathrm{s} \) in the \( y \)-direction. The wind is blowing at \( 20.0 \mathrm{~m} / \mathrm{s} \) in the \( x \)-direction. What is the direction of the velocity of the plane? \[ \theta=[?]^{\circ} \]

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