$A$ man crosses a $320 \ m$ wide river perpendicular to the current in $4 \ minutes$. If in still water he can swim with a speed $5/3$ times that of the current,then the speed of the current,in $m/min$ is

$A$ river $200 \,m$ wide is flowing at a rate of $3.0 \,m/s$. $A$ boat is sailing at a velocity of $15 \,m/s$ with respect to the water in a direction perpendicular to the river. How far from the point directly opposite to the starting point does the boat reach on the opposite bank (in $\,m$)?

$A$ boat crosses a river from one bank $A$ to another bank $B$ which is opposite. The distance between them is $D$. The speed of water is $V_W$ and that of the boat relative to water is $V_B$. If $V_B = 2V_W$, the time taken by the boat to cross the river directly along $AB$ is $(\sin 30^\circ = 1/2, \cos 30^\circ = \sqrt{3}/2)$.

The figure shows two ships moving in the $x-y$ plane with velocities $V_A$ and $V_B$. The ships move such that $B$ always remains north of $A$. The ratio $\frac{V_A}{V_B}$ is equal to ........

$A$ 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:

$A$ swimmer wants to cross a river from point $A$ to point $B$. Line $AB$ makes an angle of $30^{\circ}$ with the flow of the river. The magnitude of the velocity of the swimmer is the same as that of the river. The angle $\theta$ with the line $AB$ should be in degrees,so that the swimmer reaches point $B$.