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Question: A man running at speed of 5 km/h, finds the rain hitting his head vertically but he has to hold the ...

A man running at speed of 5 km/h, finds the rain hitting his head vertically but he has to hold the umbrella at 30° with vertical while at rest. The speed of rain w.r.t. ground is

A

53\sqrt{3} km/h

B

10 km/h

C

5 km/h

D

103\sqrt{3} km/h

Answer

10 km/h

Explanation

Solution

The problem involves relative motion, specifically the velocity of rain with respect to a moving observer. We use vector addition/subtraction.

Let:

  • v⃗R\vec{v}_R be the velocity of rain with respect to the ground.
  • v⃗M\vec{v}_M be the velocity of the man with respect to the ground.
  • v⃗R/M\vec{v}_{R/M} be the velocity of rain with respect to the man.

The fundamental relationship is: v⃗R/M=v⃗R−v⃗M\vec{v}_{R/M} = \vec{v}_R - \vec{v}_M

Let's define a coordinate system:

  • The positive x-axis is in the direction the man runs.
  • The positive y-axis is vertically upwards.

Let the speed of rain be vRv_R. Let its horizontal component be vRxv_{Rx} and its vertical component be vRyv_{Ry}. So, v⃗R=vRxi^+vRyj^\vec{v}_R = v_{Rx} \hat{i} + v_{Ry} \hat{j}. Since rain falls downwards, vRyv_{Ry} will be negative.

Scenario 1: Man is at rest. When the man is at rest, v⃗M=0\vec{v}_M = 0. The problem states that he has to hold the umbrella at 30° with the vertical. This implies that the actual velocity of rain v⃗R\vec{v}_R makes an angle of 30° with the vertical. From the vector components: The horizontal component of rain velocity is vRx=vRsin⁡30∘v_{Rx} = v_R \sin 30^\circ. The vertical component of rain velocity is vRy=−vRcos⁡30∘v_{Ry} = -v_R \cos 30^\circ (negative sign indicates downward direction).

Scenario 2: Man is running at a speed of 5 km/h. The man runs horizontally, so v⃗M=5i^\vec{v}_M = 5 \hat{i} km/h. He finds the rain hitting his head vertically. This means the relative velocity of rain with respect to the man, v⃗R/M\vec{v}_{R/M}, has no horizontal component. Using the relative velocity equation: v⃗R/M=v⃗R−v⃗M=(vRxi^+vRyj^)−5i^\vec{v}_{R/M} = \vec{v}_R - \vec{v}_M = (v_{Rx} \hat{i} + v_{Ry} \hat{j}) - 5 \hat{i} v⃗R/M=(vRx−5)i^+vRyj^\vec{v}_{R/M} = (v_{Rx} - 5) \hat{i} + v_{Ry} \hat{j}

Since the rain appears to fall vertically, the horizontal component of v⃗R/M\vec{v}_{R/M} must be zero: vRx−5=0v_{Rx} - 5 = 0 vRx=5 km/hv_{Rx} = 5 \text{ km/h}

Combining the results from both scenarios: We have two expressions for vRxv_{Rx}:

  1. vRx=vRsin⁡30∘v_{Rx} = v_R \sin 30^\circ (from Scenario 1)
  2. vRx=5 km/hv_{Rx} = 5 \text{ km/h} (from Scenario 2)

Equating these two expressions: vRsin⁡30∘=5v_R \sin 30^\circ = 5 We know sin⁡30∘=12\sin 30^\circ = \frac{1}{2}. vR×12=5v_R \times \frac{1}{2} = 5 vR=10 km/hv_R = 10 \text{ km/h}

Thus, the speed of rain with respect to the ground is 10 km/h.