A particle is projected vertically up with velocity `v=sqrt((4gR_e)/3)` from earth surface.The velocity of particle at height equal to half of the maximum height reached by it

Options

A. `sqrt((gR_(e))/2)`
B. `sqrt((gR_(e))/3)`
C. `sqrt(gR_(e))`
D. `sqrt((2gR_(e))/3)`

Answer

To solve the problem, we need to find the velocity of a particle projected vertically upwards from the Earth's surface when it reaches a height equal to half of its maximum height.

### Step-by-Step Solution:

1. **Determine the maximum height (H) reached by the particle:**
The initial velocity \( v_0 \) of the particle is given as:
\[ v_0 = \sqrt{\frac{4gR_e}{3}} \]
The maximum height \( H \) can be found using the formula:
\[ H = \frac{v_0^2}{2g} \]
Substituting the value of \( v_0 \):
\[ H = \frac{\left(\sqrt{\frac{4gR_e}{3}}\right)^2}{2g} = \frac{\frac{4gR_e}{3}}{2g} = \frac{2R_e}{3} \]

2. **Calculate the height at which we need to find the velocity:**
We need to find the velocity at a height equal to half of the maximum height:
\[ h = \frac{H}{2} = \frac{1}{2} \cdot \frac{2R_e}{3} = \frac{R_e}{3} \]

3. **Use the conservation of energy principle:**
The total mechanical energy at the Earth's surface (initial) is equal to the total mechanical energy at height \( h \):
\[ \frac{1}{2} m v_0^2 = mgh + \frac{1}{2} mv^2 \]
Here, \( v \) is the velocity at height \( h \). Rearranging gives:
\[ \frac{1}{2} m v_0^2 - mgh = \frac{1}{2} mv^2 \]
Dividing through by \( m \) and rearranging:
\[ v^2 = v_0^2 - 2gh \]

4. **Substituting the values:**
Substitute \( v_0^2 \) and \( h \):
\[ v^2 = \frac{4gR_e}{3} - 2g \cdot \frac{R_e}{3} \]
Simplifying:
\[ v^2 = \frac{4gR_e}{3} - \frac{2gR_e}{3} = \frac{2gR_e}{3} \]

5. **Finding the velocity (v):**
Taking the square root to find \( v \):
\[ v = \sqrt{\frac{2gR_e}{3}} \]

### Final Answer:
The velocity of the particle at a height equal to half of the maximum height reached by it is:
\[ v = \sqrt{\frac{2gR_e}{3}} \]