We can break this problem into two parts. We have two frames of reference: the Earth (rest frame) and the car (moving frame). The raindrops are falling vertically in the Earth frame, but to an observer in the car moving east, the raindrops appear to be coming at an angle.

A. Speed of the raindrops with respect to the car

Let’s denote:

• $v_{r/c}$: speed of raindrops relative to the car (what we need to find).
• $v_{r/e}$: speed of the raindrops relative to the Earth (falling vertically).
• $v_{c/e} = 15.0 \, \text{m/s}$: speed of the car relative to the Earth (due east).

The angle of 59.0 degrees is the angle between the vertical and the apparent path of the raindrops relative to the car. This means that the horizontal component of the raindrop’s velocity (relative to the car) is equal to the car’s velocity in the eastward direction.

Using trigonometry:

1. The horizontal component of the rain's velocity as seen from the car's frame is the car's speed, $v_{c/e}$, so: $v_{r/c} \sin(59^\circ) = 15.0 \, \text{m/s}$

2. The total speed of the raindrops relative to the car, $v_{r/c}$, can be found by rearranging the equation: $v_{r/c} = \frac{15.0}{\sin(59^\circ)}$

Let’s calculate this value:

$v_{r/c} = \frac{15.0}{\sin(59^\circ)} = \frac{15.0}{0.8572} \approx 17.5 \, \text{m/s}$

Thus, the speed of the raindrops with respect to the car is approximately $17.5 \, \text{m/s}$.

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B. Speed of the raindrops with respect to the Earth

Now, we know that the raindrops are falling vertically in the Earth’s frame, so their horizontal velocity is zero in the Earth frame. The car's horizontal velocity relative to the Earth causes the apparent angle in the car's frame.

We already calculated the relative speed of the raindrops from the car's frame, and now we want to find their vertical speed (with respect to the Earth). Using the fact that the vertical component of the raindrop's velocity is $v_{r/e}$, and the apparent angle of 59.0 degrees:

1. The vertical component of the raindrops' velocity is the adjacent side of the triangle: $v_{r/e} = v_{r/c} \cos(59^\circ)$

2. Substituting $v_{r/c} = 17.5 \, \text{m/s}$: $v_{r/e} = 17.5 \times \cos(59^\circ) = 17.5 \times 0.5150 \approx 9.0 \, \text{m/s}$

Thus, the speed of the raindrops with respect to the Earth is approximately $9.0 \, \text{m/s}$ in the vertical direction.

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Summary:

• A. The speed of the raindrops with respect to the car is approximately $17.5 \, \text{m/s}$.
• B. The speed of the raindrops with respect to the Earth is approximately $9.0 \, \text{m/s}$.

Would you like further clarification on any part of this solution, or do you need more details?

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Here are five related questions for further exploration:

1. What would happen to the angle if the car increased its speed to 25 m/s?
2. How would the answer change if the rain was falling at an angle instead of vertically?
3. What is the time it would take for the raindrops to fall a certain distance in the Earth’s frame?
4. If Oschon suddenly stopped the car, what would the new angle of the rain’s path be relative to the car?
5. How can this problem be modeled using vector addition?

Tip: When dealing with relative velocities, always consider using vector components to break down motion into perpendicular directions.