Question

Rain, pouring down at an angle \( \alpha \) with the vertical, has a constant speed of 10 m/s. A woman runs against the rain with a speed of 8 m/s and sees that the rain makes an angle \( \beta \) with the vertical. The relation between \( \alpha \) and \( \beta \) is given by:

• A. \( \tan \beta = \frac{8 + 10 \sin \alpha}{10 + 8 \cos \alpha} \)
• B. \( \tan \beta = \frac{8 \cos \alpha}{10 + 8 \sin \alpha} \)
• C. \( \tan \beta = \frac{8 + 10 \cos \alpha}{10 \sin \alpha} \)
• D. \( \tan \beta = \frac{8 + 10 \sin \alpha}{10 \cos \alpha} \)

Hint:

Relative velocity is useful in solving problems involving motion of objects against each other, like this case with rain and a moving woman.

Answer 1

The Correct Option is A

Step 1: Resolve the rain's velocity (10 m/s at angle \(\alpha\) to the vertical) into components: vertical \(10\cos\alpha\), horizontal \(10\sin\alpha\).

Step 2: The woman runs at 8 m/s horizontally, against the rain's horizontal drift, so relative to her the horizontal component becomes \(10\sin\alpha + 8\), while the vertical component stays \(10\cos\alpha\) since her velocity has no vertical part.

Step 3: The angle \(\beta\) the rain now appears to make with the vertical is \(\tan\beta = \dfrac{\text{horizontal}}{\text{vertical}}\).
\[ \boxed{\tan\beta = \dfrac{8+10\sin\alpha}{10\cos\alpha}} \]

Answer 2

Check the relation using two limiting cases. When \(\alpha=0\), the rain falls straight down, so the only horizontal motion the woman sees is her own speed of 8 m/s against a vertical component of 10 m/s, giving \(\tan\beta = 8/10\).

When \(\alpha=90^\circ\), the rain travels entirely horizontally at 10 m/s, and the woman's speed of 8 m/s adds directly to it since she runs against the rain's horizontal drift, while the vertical component of the rain vanishes, sending \(\beta\) toward \(90^\circ\). The formula \(\tan\beta = \dfrac{8+10\sin\alpha}{10\cos\alpha}\) reproduces both checks: at \(\alpha=0\) it gives \(8/10\), and as \(\alpha\to 90^\circ\) the denominator \(10\cos\alpha\to 0\) while the numerator stays positive, sending \(\tan\beta\to\infty\). \[\tan\beta = \frac{8+10\sin\alpha}{10\cos\alpha}\]