$A$ man is running at a speed of $5 \, m/s$. The rain drops appear to be falling at an angle of $45^{\circ}$ from the vertical. If the rain drops are actually falling vertically downwards,then the velocity of the rain drops (in $m/s$) is:

$A$ swimmer swims in still water at a speed of $5 \text{ km/hr}$. He enters a $200 \text{ m}$ wide river,having a river flow speed of $4 \text{ km/hr}$,at point $A$ and proceeds to swim at an angle of $127^{\circ}$ $(\sin 37^{\circ} = 0.6)$ with the river flow direction. Another point $B$ is located directly across $A$ on the other side. The swimmer lands on the other bank at a point $C$,from which he walks the distance $CB$ with a speed of $3 \text{ km/hr}$. The total time in which he reaches from $A$ to $B$ is .......... $\text{minutes}$.

Ship $A$ is moving westwards with a speed of $20 \ km h^{-1}$ and another ship $B$ which is at $200 \ km$ south of $A$ is moving northwards with a speed of $10 \ km h^{-1}$. The time after which the distance between them is shortest and the shortest distance between them are respectively

If the velocities of object $A$ and $B$ are $v_A$ and $v_B$ respectively,then write the equation of the relative velocity of $A$ with respect to $B$.

$A$ slider block $A$ moves downward at a speed of $v_A = 2 \ m/s$,at an angle of $75^\circ$ with the horizontal as shown in the figure. The velocity of the portion of belt $B$ between ideal pulleys $C$ and $D$ with respect to $A$ is $v_{CD/A} = 2 \ m/s$ at an angle $\theta$ with the horizontal. The magnitude of the velocity of portion $CD$ of the belt when $\theta = 15^\circ$ is .......... $m/s$.

$A$ man walking along a straight line with a velocity $6 \text{ km/h}$ encounters rain falling vertically down with a velocity $6\sqrt{3} \text{ km/h}$. At what angle should the man hold his umbrella to protect himself from the rain?