To swim across the river and reach a point directly opposite his starting point, the man must swim at an angle against the flow of the river to counteract the river's current.

### Given:
- Speed of the man in still water = 2 km/h
- Speed of the river flow = 1 km/h

Let \(\theta\) be the angle between the man's swimming direction and the river flow. To counteract the river's flow, the horizontal component of the man's velocity must equal the river's velocity.

\[
\text{Horizontal component} = 2 \sin \theta = 1
\]

Solving for \(\theta\):

\[
\sin \theta = \frac{1}{2}
\]
\[
\theta = \sin^{-1} \left( \frac{1}{2} \right) = 30^\circ
\]

Thus, the man should swim at an angle of:

\[
{120^\circ} \text{ upstream from the river flow.}
\]

To cross the river along the minimum distance (i.e., directly perpendicular to the riverbank), the boatman must row with a velocity component equal and opposite to the river flow to cancel the drift.

Given:
- Speed of the boat in still water = 10 km/h
- Speed of the river flow = 5 km/h
- Width of the river = 2 km

The boat's velocity perpendicular to the riverbank is:

\[
v_{\text{perpendicular}} = \sqrt{(10^2 - 5^2)} = \sqrt{100 - 25} = \sqrt{75} = 5\sqrt{3} \, \text{km/h}
\]

Time to cross the river:

\[
\text{Time} = \frac{\text{Distance}}{\text{Speed perpendicular}} = \frac{2}{5\sqrt{3}} \, \text{hours}
\]

Simplifying:

\[{\frac{2}{5\sqrt{3}} \, \text{hours}}
\]

The relative velocity of the bird with respect to the observer in the train is found by combining the bird's velocity (south) and the train's velocity (east) using the Pythagorean theorem.

\[
v_{\text{relative}} = \sqrt{(40^2 + 40^2)} = \sqrt{3200} = 40\sqrt{2} \, \text{km/h}
\]

The direction is \( 45^\circ \) south of west.

Thus, the relative velocity is:

\[ {40\sqrt{2} \, \text{km/h} \text{ at } 45^\circ \text{ south of west}}
\]

To find the velocity of the monkey as observed by a man standing on the ground, we need to add the velocity of the monkey relative to the train to the velocity of the train.

 Given:
- Velocity of the train (north direction) = 54 km/h
- Velocity of the monkey relative to the train (opposite to train's motion) = 18 km/h

Monkey's velocity relative to the ground:
Since the monkey is running against the train’s motion, the monkey's velocity relative to the ground will be:

\[
\vec{v}_{\text{monkey}} = \vec{v}_{\text{train}} - \vec{v}_{\text{monkey relative to train}}
\]

\[
v_{\text{monkey}} = 54 \, \text{km/h} - 18 \, \text{km/h} = 36 \, \text{km/h}
\]

Thus, the velocity of the monkey as observed by a man on the ground is:

\[
{36 \, \text{km/h} \text{ north} or 10 m/s \text{ north}}
\]

\[ tan \alpha = \frac{-gt}{u} \]

\[ tan (-60^{\circ })= \frac{-10t}{20/\sqrt{3}} \]

Solving t =2 sec

The tube should be inclined at an angle \(\theta\) such that the water's relative velocity to the boy is along the axis of the tube.

The horizontal velocity of the boy is \(v\), and the vertical velocity of the falling water is \(u\). The angle \(\theta\) between the tube and the vertical satisfies:

\[
\tan \theta = \frac{v}{u}
\]

Thus, the required angle is:

\[
\theta = \tan^{-1} \left( \frac{v}{u} \right)
\]

The relative velocity of the boat with respect to the water is given by subtracting the velocity of the water from the velocity of the boat.

\[
\vec{v}_{\text{bw}} = \vec{v}_{\text{boat}} - \vec{v}_{\text{water}}
\]

Given:
\[
\vec{v}_{\text{boat}} = 3\hat{i} + 4\hat{j}, \quad \vec{v}_{\text{water}} = -3\hat{i} - 4\hat{j}
\]

Now, subtract:

\[
\vec{v}_{\text{bw}} = (3\hat{i} + 4\hat{j}) - (-3\hat{i} - 4\hat{j})
\]
\[
\vec{v}_{\text{bw}} = 3\hat{i} + 4\hat{j} + 3\hat{i} + 4\hat{j}
\]
\[
\vec{v}_{\text{bw}} = 6\hat{i} + 8\hat{j}
\]

Thus, the relative velocity of the boat with respect to the water is:

\[ {6\hat{i} + 8\hat{j}} \]

To determine when the four persons \( P, Q, R, \) and \( S \) will meet, consider the following:

Relative Velocity:
- Each person moves with speed \( u \) towards the center of the square.

Configuration:
- As they face each other and move towards the center, their paths converge.

Effective Speed Towards Each Other:
- The effective speed of each person towards the center is \( u \cos 45^\circ = \frac{u}{\sqrt{2}} \) because they move diagonally.

Distance to Center:
- The distance from each person to the center of the square is \( \frac{d}{\sqrt{2}} \).

Time to Meet:
\[
t = \frac{\text{Distance}}{\text{Effective Speed}} = \frac{\frac{d}{\sqrt{2}}}{\frac{u}{\sqrt{2}}} = \frac{d}{u}
\]

Thus, the time after which they will all meet is \( \frac{d}{u} \).

As maximum height is same is all three they will have same vertical speed.

Range of C is maximum so its horizontal speed is maximum.

Given the relationship \( R = 5T^2 \), and using the formulas for range \( R \) and time of flight \( T \):

1. **Range formula:**
R = u cos(θ) · T

2. **Time of flight formula:**
T = (2u sin(θ)) / g

Substituting T into the range equation:

R = u cos(θ) · (2u sin(θ) / g)

This gives:

R = (2u² sin(θ) cos(θ)) / g

Setting this equal to \( 5T^2 \):

(2u² sin(θ) cos(θ)) / g = 5((2u sin(θ)) / g)²

Simplifying:

(2u² sin(θ) cos(θ)) / g = (20u² sin²(θ)) / g²

Cancelling \( u² / g \):

2 cos(θ) = 20 sin(θ) / g

Given g = 10 m/s²:

2 cos(θ) = 2 sin(θ)

Thus:

cos(θ) = sin(θ)

This implies:

tan(θ) = 1 → θ = 45°

The angle to the horizontal should be  45°