Given:

\[
y = 8t - 5t^2 \quad \text{and} \quad x = 6t
\]

To find the initial velocity, we calculate the horizontal and vertical components of velocity:

1. Horizontal velocity \( v_x = \frac{dx}{dt} = 6 \, \text{m/s} \)
2. Vertical velocity \( v_y = \frac{dy}{dt} = 8 - 10t \)

At \( t = 0 \) (initial velocity):

\[
v_y(0) = 8 \, \text{m/s}
\]

Thus, the magnitude of the initial velocity \( u \) is:

\[
u = \sqrt{v_x^2 + v_y^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, \text{m/s}
\]

The velocity of projection is \( 10 \, \text{m/s} \).

We Know that,

\[ \frac{R}{H}=\frac{4}{tan\theta}\]

\[ tan \theta = 4/3 \]

\[ H = \frac{u^{2}sin^{2}\theta}{2g} \]

Solving we get

\[ u= 5\sqrt{g/2}\]

The equation of motion of the projectile is given as:

\[
y = 12x - \frac{3}{4}x^2
\]

At the range, \( y = 0 \). To find the range, set \( y = 0 \) and solve for \( x \):

\[
0 = 12x - \frac{3}{4}x^2
\]

Multiply the entire equation by 4 to eliminate the fraction:

\[
0 = 48x - 3x^2
\]

Factor the equation:

\[
0 = x(48 - 3x)
\]

This gives two solutions:

\[
x = 0 \quad \text{or} \quad 48 - 3x = 0
\]

Solving for \( x \):

\[
x = \frac{48}{3} = 16
\]

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

Initial kinetic energy \( K \) is given by:

\[
K = \frac{1}{2} m u^2
\]

At the highest point, only the horizontal component of the velocity \( u \cos 60^\circ \) remains. Therefore, the kinetic energy at the highest point is due to this horizontal velocity:

\[
K_{\text{highest}} = \frac{1}{2} m (u \cos 60^\circ)^2
\]

Since \( \cos 60^\circ = \frac{1}{2} \):

\[
K_{\text{highest}} = \frac{1}{2} m \left(\frac{u}{2}\right)^2 = \frac{1}{4} \times \frac{1}{2} m u^2 = \frac{K}{2}
\]

Thus, the kinetic energy at the highest point is \( \frac{K}{2} \).

During Projectile motion horizontal component of velocity remains same so,

\[ u cos\alpha = v cos \beta \]

\[ \frac{u cos\alpha}{cos \beta} = v  \]

\[ {u .cos\alpha}.{sec \beta} = v \]

For Complementary angles ranges are equal

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

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 \( 0^\circ \).

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