To solve the problem, we need to determine the direction from which the rain droplets appear to be coming to a cyclist moving northward while the rain is falling vertically downward. We can use vector addition to find the apparent direction of the rain relative to the cyclist.
### Step-by-Step Solution:
1. **Identify the velocities:**
- The velocity of the rain droplets, \( V_r \), is \( 5 \, \text{m/s} \) downward.
- The velocity of the cyclist, \( V_c \), is \( 10 \, \text{m/s} \) northward.
2. **Set up a coordinate system:**
- Let the downward direction be negative \( y \) (i.e., \( -y \)).
- Let the northward direction be positive \( x \) (i.e., \( +x \)).
- Thus, the rain's velocity vector can be represented as \( \vec{V_r} = (0, -5) \) m/s.
- The cyclist's velocity vector can be represented as \( \vec{V_c} = (10, 0) \) m/s.
3. **Determine the relative velocity of the rain with respect to the cyclist:**
- The relative velocity \( \vec{V_{rel}} \) is given by:
\[
\vec{V_{rel}} = \vec{V_r} - \vec{V_c} = (0, -5) - (10, 0) = (-10, -5) \, \text{m/s}
\]
4. **Calculate the angle of the apparent direction of the rain:**
- The components of the relative velocity are \( V_{rel,x} = -10 \, \text{m/s} \) (westward) and \( V_{rel,y} = -5 \, \text{m/s} \) (downward).
- To find the angle \( \theta \) that the rain appears to come from, we can use the tangent function:
\[
\tan(\theta) = \frac{|V_{rel,y}|}{|V_{rel,x}|} = \frac{5}{10} = \frac{1}{2}
\]
- Therefore,
\[
\theta = \tan^{-1}\left(\frac{1}{2}\right)
\]
5. **Determine the direction:**
- The angle \( \theta \) is measured from the negative x-axis (westward) towards the negative y-axis (downward).
- This means the rain appears to come from an angle \( \theta \) above the horizontal line pointing north.
6. **Final Answer:**
- The rain droplets will appear to the cyclist to be coming from an angle \( \theta = \tan^{-1}\left(\frac{1}{2}\right) \) above the northward direction.
