Two particles are projected vertically upwards from the surface of the earth with velocities `upsilon_(1) = sqrt((2gR)/3) and upsilon_(2) = sqrt((4gR)/(3))` respectively. If the maximum heights attained by the two particles are `h_(1)` and `h_(2)` respectively, then calculate the ratio `(h_1)/(h_2)`.

Answer

To solve the problem, we will use the principle of conservation of energy. The maximum height attained by a particle projected vertically upwards can be derived from the initial kinetic energy and the change in gravitational potential energy.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- For particle 1: \( v_1 = \sqrt{\frac{2gR}{3}} \)
- For particle 2: \( v_2 = \sqrt{\frac{4gR}{3}} \)

2. **Use the Conservation of Energy:**
The total mechanical energy (kinetic + potential) at the point of projection will equal the total mechanical energy at the maximum height.

For a particle projected upwards:
\[ \text{Initial Kinetic Energy} = \text{Final Potential Energy} - \text{Initial Potential Energy} \]
\[ \frac{1}{2} mv^2 = mgh + \left(-\frac{GMm}{R}\right) \]

3. **Calculate Maximum Height for Particle 1:**
\[ \frac{1}{2} m v_1^2 = mg h_1 + \left(-\frac{GMm}{R}\right) \]
Substituting \( v_1 \):
\[ \frac{1}{2} m \left(\sqrt{\frac{2gR}{3}}\right)^2 = mg h_1 - \frac{GMm}{R} \]
Simplifying:
\[ \frac{1}{2} m \cdot \frac{2gR}{3} = mg h_1 - \frac{GMm}{R} \]
\[ \frac{mgR}{3} = mg h_1 - \frac{GMm}{R} \]
Rearranging gives:
\[ mg h_1 = \frac{mgR}{3} + \frac{GMm}{R} \]
Dividing by \( mg \):
\[ h_1 = \frac{R}{3} + \frac{GM}{gR} \]

4. **Calculate Maximum Height for Particle 2:**
\[ \frac{1}{2} m v_2^2 = mg h_2 + \left(-\frac{GMm}{R}\right) \]
Substituting \( v_2 \):
\[ \frac{1}{2} m \left(\sqrt{\frac{4gR}{3}}\right)^2 = mg h_2 - \frac{GMm}{R} \]
Simplifying:
\[ \frac{1}{2} m \cdot \frac{4gR}{3} = mg h_2 - \frac{GMm}{R} \]
\[ \frac{2mgR}{3} = mg h_2 - \frac{GMm}{R} \]
Rearranging gives:
\[ mg h_2 = \frac{2mgR}{3} + \frac{GMm}{R} \]
Dividing by \( mg \):
\[ h_2 = \frac{2R}{3} + \frac{GM}{gR} \]

5. **Calculate the Ratio \( \frac{h_1}{h_2} \):**
\[ \frac{h_1}{h_2} = \frac{\left(\frac{R}{3} + \frac{GM}{gR}\right)}{\left(\frac{2R}{3} + \frac{GM}{gR}\right)} \]
Let \( x = \frac{GM}{gR} \):
\[ \frac{h_1}{h_2} = \frac{\frac{1}{3} + x}{\frac{2}{3} + x} \]
Cross-multiplying gives:
\[ = \frac{1 + 3x}{2 + 3x} \]
As \( x \) approaches zero (for large R), this simplifies to:
\[ \frac{1}{2} \]

6. **Final Result:**
Therefore, the ratio of the maximum heights attained by the two particles is:
\[ \frac{h_1}{h_2} = \frac{1}{2} \]