Rain is falling vertically with a velocity of 4kmhr^(-1) . A cyclist is going along a horizontal road with a velocity of 3km hr^(-1) towards east. 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?

Rain is falling vertically with a velocity of 4kmhr^(-1) . A cyclist i

1.	Rain is falling vertically with a velocity of 4kmhr^(-1) . A cyclist is going along a horizontal road with a velocity of 3km hr^(-1) towards east. 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?
Answer» Solution :The SITUATION is shown in Fig. OA represents velocity `vecv_(c) ` of cyclist , which is `3km hr^(-1)` OB represents velocity `vecv_(R)` of vertically falling rain , which is `4km 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=3km hr^(-1), OB=4km hr^(-1)` `/_BOC=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).