A stationary person observes that rain is falling vertically down at `30 km h^-1`. A cyclist is moving up on an inclined plane making an angle `30^@` with horizontal at `10 km h^-1`. In which direction should the cyclist hold his umbrella to prevent himself from the rain ?

Options

A. At an angle`tan^-1 ((3sqrt(3))/(5))` with inclined plane.
B. At an angle `tan^-1 ((3sqrt(3))/(5))` with horizontal.
C. A an angle `tan^-1 (sqrt(3)/(7))` with inclined plane.
D. At an angle `tan^-1 (sqrt(3)/(7))` with vertical.

Answer

A, D

Explanation

(a.,d.) `vec v_r = -30 hat j, vec v_m = (10 sqrt(3))/(2) hat i+ (10)/(2) hat j`
`tan theta = (5 sqrt(3))/(35) = (sqrt(3))/(7) rArr theta = tan^-1 ((sqrt(3))/(7))` with vertical
Angle of `v_(r//m)` with inclined plane is
`60 - theta = beta`
`tan beta = tan(60 - theta)`
=`(tan 60^@ - tan theta)/(1 + tan 60^@ tan theta)`
=`(sqrt(3) - sqrt(3) //7)/(1 + sqrt(3). sqrt(3//7)) =(3 sqrt(3))/(5)`
`rArr beta = tan^-1((3 sqrt(3))/(5))`.
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