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 \( {6.0 \, \text{m/s}} \).

1. Relative Speed (Same Direction):
\[
\text{Speed of Train} = 15 \, \text{m/s}
\]
\[
\text{Time to Overtake} = 10 \, \text{s}
\]
\[
\text{Distance} = \text{Length of Train} = 100 \, \text{m}
\]
\[
\text{Relative Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{100}{10} = 10 \, \text{m/s}
\]
\[
\text{Speed of Man} = 15 - 10 = 5 \, \text{m/s}
\]

2. Relative Speed (Opposite Direction):
\[
\text{Relative Speed} = 15 + 5 = 20 \, \text{m/s}
\]

3. Time to Cross Man (Opposite Direction):
\[
\text{Time} = \frac{\text{Distance}}{\text{Relative Speed}} = \frac{100}{20} = 5 \, \text{s}
\]

Thus, the train will take 5 seconds to cross the man if he is running in the opposite direction.

Horizontal Displacement = πR

Vertical Displacement = 2R

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

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

When particle comes back to initial position displacement becomes zero.

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 \( 0^\circ \).

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

As A is having more maximum height it will have more vertical speed u_y. So A will take more time than A.

So, ship B will be hit first.

Distance is equal for both A and B. B is taking lesser time to complete so its speed is higher.