A person standing on a road has to hold his umbrella at `60^(0)` with the verticcal to keep the rain away. He throws the umbrella an starts running at `20 ms^(-1)`. He finds that rain drops are hitting his head vertically. Find the speed of the rain drops with respect to (a) the road (b) the moving person.

Explanation

Let velocity of rain w.r.t gorund is ` vecv_() = xhati - yhatj`
Case - I
then ` vecv_(R.b) = vecv_(R) -vecV_(b) = x hati -yhatj -0 = xhati -yhatj`
` tan 60^(@) = (|xhatj|)/(|-yhatj|) =- x/y =sqrt3`
` x = sqrt3y`
case - II
` vecv _(R.b) = vecv_(R) -vecv_(b)`
= ` xhati -hatj -20hati`
` = (x -20)hati -yhatj`
when velcoity of rain w.r.t boy si vertically downward then ` v_(x)` i.e, coefficient of ` hati` is zero.
x -20=0
x=20
From eq (i)
` x = sqrt y = 20 Rightarrow y= 20/sqrt3`
Then ` v_(R) = 20 hati = 20/sqrt3 hatj`
` |vecv_(R)| = sqrt((20)^(2) + (- 20/sqrt3)^(2)) = sqrt(400 + 400/3) = sqrt(1600/3 = 40/sqrt3`
` v_(R) = 40/sqrt3 m//s`
` vecv_(R.b) = vecv_(R)-vecv_(b) = 20hati - 20/sqrt3 hatj - 20hati - 20/sqrt3 hatj`
`|vec_(R.b)|20/sqrt3 ` m/s