Rain is falling vertically with a velocity of 4 km `hr^(-1)` . A cyclist is going along a horizontal road with a velocity of 3 km `hr^(-1)` towards cast. Calculate the relative velocity of the rain with respect to the cyclist. In what direction must the cyclist hold his umbrella to save himself from the falling rain ?

Explanation

The situation is shown in Fig.
OA represents velocity `vecV_(C ) `of cyclist, which is 3 km `hr^(-1)`
OB represents velocity `vecC_(R)` of vertically falling rain, which is 4 km `hr^(-1)`
OC represents `-vecv_(C )` opposite of velocity of the cyclist.
OD represents `vecV_R +(-vecv_C )= vecv_R - vecv_C ` velocity of rain relative to cyclist.
In parallelogram OBDC,
`OC = 3 km hr^(-1) OB = 4 km hr^(-1)`
`angleBOC = 90^@ `
Hence `OD = sqrt((OC)^2 +(OB)^2 )= sqrt( (3)^2 +(4)^2) = sqrt(9+ 16 ) = sqrt(25) = 5 km hr^(-1)`
`tan Beta =(BD )/(OB )= (OC )/(OB ) = 3/4 = 0.75 = tan 36^@ 52 ., i.e beta =36^@ -52.` east of vertical .
The cyclist must hold his umbrella at `36^@ - 52.` with vertical in the direction of his motion (i.e. east).