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 \( 1 \, \text{kg m/s} \).

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.

At the highest point of its trajectory, the vertical component of the velocity becomes zero, while the horizontal component remains unchanged.

- Initial velocity: \( u \)
- Horizontal component: \( u_x = u \cos \theta \)
- Vertical component: \( u_y = u \sin \theta \)

At the highest point:
- Vertical velocity: \( u_y = 0 \)
- Horizontal velocity: \( u_x = u \cos \theta \)

The change in velocity is only in the vertical direction, from \( u \sin \theta \) to 0.

So, the change in velocity is:

\[
\Delta v = u \sin \theta
\]

Thus, the change in velocity is \( u \sin \theta \).

Maximum height \(H = \frac{u^2\sin^2\theta}{2g}\) and range \(R = \frac{2u^2\sin\theta\cos\theta}{g}\). Equating the two yields \(\frac{\sin^2\theta}{2} = 2\sin\theta\cos\theta\), which simplifies to \(\tan\theta = 4\).

We require \(H = \frac{R}{2}\). Using the formulae \(H = \frac{u^2 \sin^2 \theta}{2g}\) and \(R = \frac{u^2 \sin 2\theta}{g}\), we get \(\frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin\theta\cos\theta}{g} ⇒ \tan\theta = 2 ⇒ \theta = \tan^{-1}(2)\).

Range is given by \(R = \frac{u^2 \sin(2\theta)}{g}\). For \(\theta = 15^\circ\), \(R_1 = \frac{u^2 \sin(30^\circ)}{g} = 1.5\text{ km}\) which implies \(\frac{u^2}{g} = 3.0\text{ km}\). For \(\theta = 45^\circ\), \(R_2 = \frac{u^2 \sin(90^\circ)}{g} = \frac{u^2}{g} = 3\text{ km}\).

Horizontal speed is \(u_x = u\cos(53^\circ) = 30\text{ m/s}\), and vertical speed is \(u_y = u\sin(53^\circ) = 40\text{ m/s}\). At \(45^\circ\) on the way down, \(v_y = -v_x = -30\text{ m/s}\). Using \(v_y = u_y - gt\), we get \(-30 = 40 - 10t\), which yields \(t = 7\text{ s}\).

Comparing the trajectory to \(y = xtan\theta - \frac{gx^2}{2u^2cos^2\theta}\), we find \(tan\theta = 1⇒ \theta = 45^\circ\). Range at \(y = 0\) is \(x = 80\text{ m}\). Maximum height is \(H = \frac{Rtan\theta}{4} = 20\text{ m}\), and the initial speed \(u = 20\sqrt{2}\text{ m/s}\). Hence, A-Q, B-P, C-R, D-S.