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To a man running at a speed of 20 m/hr, ...

To a man running at a speed of 20 m/hr, the rain drops appear to be falling at an angle of `30^(@)` from the vertical. If the rain drops are actually falling vertically downwards, their velocity in km.hr is

A

`10 sqrt"" 3`

B

10

C

`20 sqrt"" 3`

D

40

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the situation using the concept of relative velocity. The man is running at a speed of 20 m/hr, and the rain appears to be falling at an angle of 30 degrees from the vertical. We need to find the actual velocity of the rain in km/hr. ### Step-by-Step Solution: 1. **Understanding the Given Information**: - Speed of the man (Vm) = 20 m/hr - Angle of apparent rain (θ) = 30 degrees - The actual velocity of the rain (Vr) is vertical. 2. **Convert the Speed of the Man**: - Since we need to find the rain's velocity in km/hr, we convert the man's speed: \[ Vm = 20 \text{ m/hr} = \frac{20}{1000} \text{ km/hr} = 0.02 \text{ km/hr} \] 3. **Setting Up the Relative Velocity**: - The velocity of the rain with respect to the man (Vr,m) can be expressed using trigonometry: \[ \tan(θ) = \frac{Vm}{Vr} \] - Here, θ = 30 degrees, and we know that: \[ \tan(30) = \frac{1}{\sqrt{3}} \] 4. **Substituting Values into the Equation**: - We can rearrange the equation to find Vr: \[ \frac{1}{\sqrt{3}} = \frac{0.02}{Vr} \] - Cross-multiplying gives us: \[ Vr = 0.02 \cdot \sqrt{3} \] 5. **Calculating the Value of Vr**: - We know that \(\sqrt{3} \approx 1.732\): \[ Vr = 0.02 \cdot 1.732 \approx 0.03464 \text{ km/hr} \] 6. **Final Conversion to km/hr**: - Since we need to express the final answer in km/hr, we can simply state: \[ Vr \approx 0.03464 \text{ km/hr} \] ### Final Answer: The actual velocity of the rain is approximately **0.03464 km/hr**.
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