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:

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

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:

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

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:

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

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.

The momentum change occurs only in the vertical direction as the horizontal component of velocity remains constant throughout the flight.

### Initial vertical component of velocity:
\[
u_y = u \sin \theta = 20 \times \sin 30^\circ = 20 \times \frac{1}{2} = 10 \, \text{m/s}
\]

At the highest point, the vertical component of velocity becomes zero.

### Change in vertical velocity:
\[
\Delta v_y = u_y - 0 = 10 \, \text{m/s}
\]

### Mass of the projectile:
\[
m = 100 \, \text{g} = 0.1 \, \text{kg}
\]

### Change in momentum:
\[
\Delta p = m \times \Delta v_y = 0.1 \times 10 = 1 \, \text{kg m/s}
\]

Thus, the change in momentum is \( 1 \, \text{kg m/s} \).

Given:

- Horizontal distance between buildings \( x = 30 \, \text{m} \)
- Height difference \( h = 150 - 27.5 = 122.5 \, \text{m} \)
- Use \( h = \frac{1}{2} g t^2 \) to find time \( t \):

\[
122.5 = \frac{1}{2} \times 10 \times t^2 \quad \Rightarrow \quad t^2 = 24.5 \quad \Rightarrow \quad t = \sqrt{24.5} \approx 4.95 \, \text{seconds}
\]

Horizontal velocity \( v = \frac{x}{t} = \frac{30}{4.95} \approx 6.06 \, \text{m/s} \).

Thus, the required speed is \( {6.0 \, \text{m/s}} \).

Horizontal Displacement = πR

Vertical Displacement = 2R

Total Displacement = ((πR)^2+(2R)^2)^1/2=

\[\sqrt{\pi^{2}+4} m\]

Slope of x-t graph represents velocity. slope is more for B so it will have higher velocity

To find the angle with the horizontal after 1.5 seconds, we calculate the horizontal and vertical components of velocity at that moment.

- Initial speed: \( u = 30 \, \text{m/s} \)
- Angle of projection: \( \theta = 30^\circ \)
- Horizontal component of velocity (remains constant):
\[
u_x = u \cos \theta = 30 \times \frac{\sqrt{3}}{2} = 15\sqrt{3} \, \text{m/s}
\]
- Vertical component of velocity after 1.5 seconds:
\[
u_y = u \sin \theta - g t = 30 \times \frac{1}{2} - 10 \times 1.5 = 15 - 15 = 0 \, \text{m/s}
\]

Since the vertical component is 0, the angle with the horizontal is:

\[
\text{Angle} = 0^\circ
\]

Thus, the angle with the horizontal after 1.5 seconds is \( 0^\circ \).

\[ tan \phi= H/2R= \frac{1}{2}tan\theta \]