Home Physics Motion in a Straight Line Relative Motion A man standing on a road hold his umbrella a…
Physics Motion in a Straight Line Relative Motion Single Correct MCQ
Published on: September 12, 2026

A man standing on a road hold his umbrella at 30º with the vertical to keep the rain away. He throws the umbrella and starts running at 10 km/hr. He finds that raindrops are hitting his head vertically, the speed of raindrops with respect to the road will be –

A
10 km/hr
B
20 km/hr
C
30 km/hr
D
40 km/hr

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Text Solution

Verified by Experts
The correct answer is:
B
Step 1: Understand the scenario. When the man holds the umbrella at an angle of 30º with the vertical, it indicates that the rain has a horizontal component of velocity that allows the rain to not hit him while he is stationary.
Step 2: When he runs at 10 km/hr, he finds the rain is hitting him vertically, which means he has compensated for the horizontal component of the rain's velocity.
Step 3: The horizontal component of the rain's velocity can be calculated using trigonometric principles. If we let $v_r$ be the vertical speed of the raindrops, then the horizontal speed of the rain is related to the angle. The tangent of the angle θ (30º) is given by:
$$ an(30^ ext{o}) = \frac{\text{horizontal component}}{\text{vertical component}} $$
Step 4: Calculate the horizontal component. The tangent of 30º is \( \frac{1}{\sqrt{3}} \), so we have:
$$ \frac{10 \text{ km/hr}}{v_r} = \frac{1}{\sqrt{3}} \implies v_r = 10\sqrt{3} \text{ km/hr} $$
Step 5: To find the actual speed of the raindrops, we need to combine the vertical and horizontal components using the Pythagorean theorem since they are perpendicular to each other. Therefore, we calculate the resultant speed of the rain as follows:
$$ v = \sqrt{(10\text{ km/hr})^2 + (10\sqrt{3}\text{ km/hr})^2} $$
$$ = \sqrt{100 + 300} = \sqrt{400} = 20 \text{ km/hr} $$
Conclusion: Thus, the speed of raindrops with respect to the road is 20 km/hr.
Therefore, the final answer is B.

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