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 wigh respect to (a) the road (b) the moving person.

Options

A. `40/3m//s,20/3m//s`
B. `40/3m//s,22/3m//s`
C. `40sqrt3/3m//s,20sqrt3/3m//s`
D. `40sqrt3/3m//s,20/3m//s`

Answer

C

Explanation

Given `theta=60^(0)` and velocity of person `vecV_(P)=vec(OA)=20ms^(-1)`.This velocity Is same as the velocity of person w.r.t ground.First of all let's see how the diagram works out.
`vecV_(r//p)=vec(OB)=`veclocity of rain w.r.t person.
`vecv_(r)=vec(OC)=`velocity of rain w.r.t earth
Values of `vecv` and `vecv_(r//p)` can be obtained by using simple trignometric relations
1.Speed of rain drops w.r.t earth `vecv=vec(OC)` from `DeltaOCB,(CB)/(OC)=sin 60^(0)`
`rArr OC=(CB)/(sin 60^(0))=(20)/(sqrt(3)//2)=40/sqrt3=(40sqrt3)/3ms^(-1)`.
2. Speed of rain w.r.t the person `vecv_(r//p)-=vecOB`
From `DeltaOCM,(OB)/(CB)=cot 60^(0)`
`rArr OB=CBcot60^(0)=20/sqrt3=(20sqrt3)/3ms^(-1)`