$A$ boy standing on the footpath tosses a ball straight up and catches it. The driver of a car passing by,moving with uniform velocity,sees this. The trajectory of the ball as seen by the driver will be:

$A$ ship is travelling due east at $10 \, km/hr$. $A$ second ship,heading $30^o$ east of north,is always due north of the first ship. The speed of the second ship in $km/hr$ is:

$A$ 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?

Rain is falling at an angle of $30^{\circ}$ from the vertical due to the wind with a speed of $40 \ m/s$. $A$ car is travelling horizontally in the direction opposite to the wind, at a speed of $40 \ m/s$. At what angle from the vertical will it experience the rain falling from (in $^{\circ}$)?

$A$ 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:

$A$ man standing on a road has to hold his umbrella at $30^{\circ}$ with the vertical to keep the rain away. He throws the umbrella and starts running at $10 \,km/h$. He finds that raindrops are hitting his head vertically. The actual speed of raindrops is: