Rain is falling with constant and uniform velocity v at angle 37º vertical. An observer starts from rest and moves with uniform acceleration in horizontal direction (with the rain). After 6 sec. velocity of rain drop again appear at angle 37º from vertical. Velocity of rain fall with respect to ground is:

JEE MainPhysicsToughRelative motion

Rain is falling with constant and uniform velocity v at angle
37º vertical. An observer starts from rest and moves with uniform
acceleration $2m/s^{2}$ in horizontal direction (with the rain). After 6
sec. velocity of rain drop again appear at angle 37º from vertical.
Velocity of rain fall with respect to ground
is: $\left( tan37^{\circ} \right.\ = 3/4)$

Show answer and solution

w.r.t observer
let final velocity of rain be v'
initial velocity of
rain

= \frac{3v}{5}\overset{\hat{}}{i} - \frac{4v}{5}\overset{\hat{}}{j}

[IMAGE] w.r.t observer.
acceleration of rain

= - 2\overset{\hat{}}{j}m/s^{2}

final velocity of
rain

= - \frac{3v^{'}}{5}\overset{\hat{}}{i} - \frac{4v^{'}}{5}\overset{\hat{}}{j}

in y- direction

V_{y\ final\ } = V_{y\ initial\ }

v^{'} = v

Source: Grass-G / P-05 Physics paper 10 October.docx

Scan any doubt from your notes and get a whiteboard explanation.