Q.10 Rain is falling vertically with a speed of 40 m s–1. Wind starts blowing with a speed
of 16 m s–1 in the west to east direction. How should a person, who is standing, hold
his umbrella to avoid getting wet?
(A) At an angle of about 22° with vertical towards east
(B) At an angle of about 22° with vertical towards west
(C) At an angle of about 66° with vertical towards east
(D) At an angle of about 66° with vertical towards west
Problem Overview
Rain falls vertically at 40 m/s, but wind blowing west to east at 16 m/s creates an effective rain velocity at an angle. The person must tilt the umbrella opposite to this direction, towards the east at about 22° from vertical.
Problem Analysis
Rain’s initial velocity is purely vertical: \(\vec{v_r} = -40\hat{j}\) m/s (downward). Wind adds horizontal velocity: \(\vec{v_w} = +16\hat{i}\) m/s (west to east, assuming east as positive x).
Resultant velocity \(\vec{v_{net}} = 16\hat{i} – 40\hat{j}\) m/s. The angle \(\theta\) with vertical follows \(\tan\theta = \frac{v_{horizontal}}{v_{vertical}} = \frac{16}{40} = 0.4\), so \(\theta = \tan^{-1}(0.4) \approx 21.8^\circ\) or about 22°.
Umbrella aligns opposite the resultant, tilting east from vertical to block slanted drops.
Option Breakdown
| Option | Angle Calculation | Direction Logic | Correct? |
|---|---|---|---|
| (A) 22° towards east | \(\tan^{-1}(16/40)\) | Opposite east drift | Yes |
| (B) 22° towards west | Correct angle, wrong side | Allows rain through | No |
| (C) 66° towards east | \(\tan^{-1}(40/16)\) | Wrong reference (horizontal) | No |
| (D) 66° towards west | Inverted ratio + side | Physically invalid | No |
Vector Solution Steps
Rain falling vertically with wind blowing west to east creates a classic relative velocity problem in physics, especially relevant for CSIR NET Life Sciences students tackling kinematics in exam prep. When rain drops at 40 m/s vertically and wind adds 16 m/s horizontally west to east, the resultant velocity slants the rain path.
Understand the physics: Vertical rain velocity \(v_r = 40\) m/s downward combines with horizontal wind \(v_w = 16\) m/s eastbound. Resultant \(v_{net}\) has magnitude \(\sqrt{40^2 + 16^2} = 42.85\) m/s, but direction matters most: \(\theta = \tan^{-1}(16/40) = 21.8^\circ \approx 22^\circ\) from vertical towards east.
To avoid getting wet, hold umbrella along \(-\vec{v_{net}}\), tilting 22° east from vertical. This blocks slanted drops effectively for a stationary person.
Common Exam Mistakes
Many confuse direction: Wind west-to-east pushes rain east, so tilt east (not west). Angle 66° errors arise from inverting ratio \(\tan^{-1}(40/16)\), measuring from horizontal instead.
Practice this for CSIR NET: Vector addition applies across biology exam kinematics questions on motion, diffusion rates, or particle tracks.[memory:1][conversation_history:5]
