$A$ man standing on a road holds his umbrella at $30^\circ$ with the vertical to keep the rain away. He throws the umbrella and starts running at $10 \ km/hr$. He finds that raindrops are hitting his head vertically. The speed of raindrops with respect to the road will be ......... $km/hr$.

$A$ river is flowing from $W$ to $E$ with a speed of $5 \, m/min$. $A$ man can swim in still water with a velocity of $10 \, m/min$. In which direction should the man swim to take the shortest possible path to reach the opposite bank (South)?

$A$ river is flowing due east with a speed $3\, ms^{-1}$. $A$ swimmer can swim in still water at a speed of $4\, ms^{-1}$ (figure).
$(a)$ If the swimmer starts swimming due north,what will be his resultant velocity (magnitude and direction)?
$(b)$ If he wants to start from point $A$ on the south bank and reach the opposite point $B$ on the north bank,
$(i)$ In which direction should he swim?
$(ii)$ What will be his resultant speed?
$(c)$ From the two different cases as mentioned in $(a)$ and $(b)$ above,in which case will he reach the opposite bank in a shorter time?

There are two particles $P$ and $Q$ separated by a distance of $10 \ km$ and moving with velocities of $10 \ kmph$ as shown in the figure. The time elapsed before they reach their minimum separation is.........$hr$.

The speed of a swimmer is $4\,km\,h^{-1}$ in still water. If the swimmer makes his strokes normal to the flow of a river of width $1\,km$,he reaches a point $750\,m$ downstream on the opposite bank. The speed of the river water is $.........\,km\,h^{-1}$.

$A$ particle $(A)$ moves due north at $3\,km/h$ and another particle $(B)$ moves due west at $4\,km/h$. The relative velocity of $A$ with respect to $B$ is $(\tan 37^{\circ} = 3/4)$.