A man standing still finds rain pouring at angle `30^(@)` with vertical , when he starts walking forward at a speed of 4Kmph it appears to him that rain pours vertically down. When he runs at triple the speed, what is the angle at which rain appers to pour ?

Explanation

Case I : When the man is still
`V_(R)` = actual (absolute )
velocity of rain.
Case : II
When man walks at 4kph.
rain fall vertically
mean horizontal component of
rains. velocity = 4
`therefore V_(R) sin 30^(@) = 4 rArr V_(R) = 8`

Case : III
The man runs at triple the speed `V_(M) = 12kph`.
we have superimposed `V_(M) ` to `V_(R) ` to
get the resultant .
From figure we see `tan phi = (O.B)/(OO.) = (12-4)/(sqrt(8^(2) - 4^(2)) = (8)/(sqrt(48)) = (2)/(sqrt(3)) therefore phi = 49`
and raindrop makes an angle of `49.6^(@)` in opposite direction.