A projectile is projected tangentially from the surface of a planet of radius R. If it is at a height of 3R at the farthest point of its trajectory, then the velocity of projection `V_(0)` is given by ( acceleration due to gravity on surface=g)

Options

A. `v_(0)=sqrt(1.5gR)`
B. `v_(0)=sqrt(0.5gR)`
C. `v_(0)=sqrt(1.6gR)`
D. `v_(0)=sqrt(2gR//3)`

Answer

To solve the problem, we will use the principle of conservation of mechanical energy. The key points to note are the initial and final states of the projectile.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A projectile is projected tangentially from the surface of a planet with radius \( R \).
- At the highest point of its trajectory, the projectile reaches a height of \( 3R \) above the surface. Therefore, the distance from the center of the planet to the projectile at the highest point is \( R + 3R = 4R \).

2. **Applying Conservation of Mechanical Energy**:
- The total mechanical energy at the surface (initial state) is the sum of the initial potential energy and initial kinetic energy.
- The total mechanical energy at the highest point (final state) is the sum of the potential energy and kinetic energy at that point.

3. **Initial Energy Calculation**:
- The initial potential energy \( U_i \) at the surface is given by:
\[ U_i = -\frac{GMm}{R} \]
- The initial kinetic energy \( K_i \) is:
\[ K_i = \frac{1}{2} m V_0^2 \]
- Therefore, the total initial energy \( E_i \) is:
\[ E_i = U_i + K_i = -\frac{GMm}{R} + \frac{1}{2} m V_0^2 \]

4. **Final Energy Calculation**:
- At the highest point (distance \( 4R \)), the potential energy \( U_f \) is:
\[ U_f = -\frac{GMm}{4R} \]
- At the highest point, the kinetic energy \( K_f \) is zero (the projectile momentarily stops before falling back).
- Therefore, the total final energy \( E_f \) is:
\[ E_f = U_f + K_f = -\frac{GMm}{4R} + 0 = -\frac{GMm}{4R} \]

5. **Setting Initial Energy Equal to Final Energy**:
- According to the conservation of energy:
\[ E_i = E_f \]
- Thus, we have:
\[ -\frac{GMm}{R} + \frac{1}{2} m V_0^2 = -\frac{GMm}{4R} \]

6. **Simplifying the Equation**:
- Cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ -\frac{GM}{R} + \frac{1}{2} V_0^2 = -\frac{GM}{4R} \]
- Rearranging gives:
\[ \frac{1}{2} V_0^2 = -\frac{GM}{4R} + \frac{GM}{R} \]
- Combine the terms on the right:
\[ \frac{1}{2} V_0^2 = \frac{GM}{R} \left(1 - \frac{1}{4}\right) = \frac{GM}{R} \cdot \frac{3}{4} \]

7. **Solving for \( V_0^2 \)**:
- Multiply both sides by 2:
\[ V_0^2 = \frac{3GM}{2R} \]

8. **Expressing in terms of \( g \)**:
- Recall that \( g = \frac{GM}{R^2} \), thus \( GM = gR^2 \):
\[ V_0^2 = \frac{3gR^2}{2R} = \frac{3gR}{2} \]
- Therefore, taking the square root gives:
\[ V_0 = \sqrt{\frac{3gR}{2}} \]

### Final Answer:
\[ V_0 = \sqrt{\frac{3gR}{2}} \]