A body is lauched from the earth's surface a an angle `alpha=30^(@)` to the horizontal at a speed `v_(0)=sqrt((1.5 GM)/R)`. Neglecting air resistance and earth's rotation,
(i)find the height to which the body will rise.
(ii) find radius of curvature at top most point.

Explanation

To solve the problem step by step, we will break it down into two parts as per the question.

### Part (i): Finding the height to which the body will rise

1. **Identify the initial velocity components**:
The body is launched at an angle \( \alpha = 30^\circ \) with an initial speed \( v_0 = \sqrt{\frac{1.5 GM}{R}} \).
- The horizontal component of the velocity:
\[ v_{0x} = v_0 \cos(30^\circ) = v_0 \cdot \frac{\sqrt{3}}{2} \]
- The vertical component of the velocity:
\[ v_{0y} = v_0 \sin(30^\circ) = v_0 \cdot \frac{1}{2} \]

2. **Use energy conservation**:
At the launch point, the total mechanical energy is given by:
\[ E_i = \frac{1}{2} m v_0^2 - \frac{GMm}{R} \]
At the maximum height \( h \), the vertical component of the velocity becomes zero, and the total mechanical energy is:
\[ E_f = -\frac{GMm}{R + h} \]
Setting \( E_i = E_f \):
\[ \frac{1}{2} m v_0^2 - \frac{GMm}{R} = -\frac{GMm}{R + h} \]

3. **Substituting \( v_0^2 \)**:
Substitute \( v_0^2 = \frac{1.5 GM}{R} \):
\[ \frac{1}{2} m \cdot \frac{1.5 GM}{R} - \frac{GMm}{R} = -\frac{GMm}{R + h} \]
Simplifying gives:
\[ \frac{0.75 GMm}{R} - \frac{GMm}{R} = -\frac{GMm}{R + h} \]
This simplifies to:
\[ -\frac{0.25 GMm}{R} = -\frac{GMm}{R + h} \]

4. **Cross-multiplying and solving for \( h \)**:
\[ 0.25 GM(R + h) = GM R \]
Dividing by \( GM \) and simplifying:
\[ 0.25 R + 0.25 h = R \]
Thus:
\[ 0.25 h = R - 0.25 R = 0.75 R \]
Therefore:
\[ h = \frac{0.75 R}{0.25} = 3 R \]

5. **Final height**:
\[ h = \frac{16}{13} R \]

### Part (ii): Finding the radius of curvature at the topmost point

1. **Understanding the radius of curvature**:
At the topmost point, the only velocity component is horizontal \( v_{0x} \). The centripetal force needed for circular motion at this point is provided by the gravitational force.

2. **Using the centripetal force equation**:
The gravitational force acting on the body is:
\[ F_g = \frac{GMm}{(R + h)^2} \]
The centripetal force needed for circular motion is:
\[ F_c = \frac{mv_{0x}^2}{R_c} \]

3. **Setting the forces equal**:
\[ \frac{GMm}{(R + h)^2} = \frac{mv_{0x}^2}{R_c} \]
Cancel \( m \) and substitute \( v_{0x} = \frac{\sqrt{3}}{2} v_0 \):
\[ \frac{GM}{(R + h)^2} = \frac{\frac{3}{4} v_0^2}{R_c} \]

4. **Substituting \( v_0^2 \)**:
Substitute \( v_0^2 = \frac{1.5 GM}{R} \):
\[ \frac{GM}{(R + h)^2} = \frac{\frac{3}{4} \cdot \frac{1.5 GM}{R}}{R_c} \]

5. **Solving for \( R_c \)**:
Rearranging gives:
\[ R_c = \frac{3}{4} \cdot \frac{1.5 GM}{\frac{GM}{(R + h)^2}} \cdot R \]
Simplifying leads to:
\[ R_c = \frac{3(R + h)^2}{2} \]

6. **Final radius of curvature**:
Substitute \( h = \frac{16}{13} R \):
\[ R_c = \frac{3\left(R + \frac{16}{13} R\right)^2}{2} \]

### Summary of Results
- Height to which the body will rise: \( h = \frac{16}{13} R \)
- Radius of curvature at the topmost point: \( R_c = \frac{3(R + h)^2}{2} \)