A particle is projected vertically upwards with a velocity `sqrt(gR)`, where `R` denotes the radius of the earth and `g` the acceleration due to gravity on the surface of the earth. Then the maximum height ascended by the particle is

Explanation

To solve the problem of finding the maximum height ascended by a particle projected vertically upwards with a velocity of \(\sqrt{gR}\), we can use the principle of conservation of energy. Here’s the step-by-step solution:

### Step 1: Write down the initial kinetic energy and potential energy.
The initial kinetic energy (KE) of the particle when it is projected is given by:
\[ KE = \frac{1}{2} mv^2 = \frac{1}{2} m (\sqrt{gR})^2 = \frac{1}{2} m (gR) = \frac{mgR}{2} \]

The initial potential energy (PE) at the surface of the Earth is:
\[ PE = -\frac{GMm}{R} \]
where \(G\) is the universal gravitational constant, \(M\) is the mass of the Earth, and \(R\) is the radius of the Earth.

### Step 2: Write down the final kinetic energy and potential energy at the maximum height.
At the maximum height \(h\), the kinetic energy becomes zero (since the particle momentarily stops before falling back down), and the potential energy is:
\[ PE_{final} = -\frac{GMm}{R + h} \]

### Step 3: Apply the conservation of energy principle.
According to the conservation of energy:
\[ \text{Initial KE} + \text{Initial PE} = \text{Final KE} + \text{Final PE} \]
Substituting the values we found:
\[ \frac{mgR}{2} - \frac{GMm}{R} = 0 - \frac{GMm}{R + h} \]

### Step 4: Simplify the equation.
First, we can cancel \(m\) from all terms (assuming \(m \neq 0\)):
\[ \frac{gR}{2} - \frac{GM}{R} = -\frac{GM}{R + h} \]

### Step 5: Substitute \(g\) in terms of \(G\) and \(M\).
We know that:
\[ g = \frac{GM}{R^2} \]
Substituting this into our equation gives:
\[ \frac{GM}{R^2} \cdot R/2 - \frac{GM}{R} = -\frac{GM}{R + h} \]
This simplifies to:
\[ \frac{GM}{2R} - \frac{GM}{R} = -\frac{GM}{R + h} \]

### Step 6: Combine the terms.
The left-hand side can be combined:
\[ \frac{GM}{2R} - \frac{2GM}{2R} = -\frac{GM}{R + h} \]
This simplifies to:
\[ -\frac{GM}{2R} = -\frac{GM}{R + h} \]

### Step 7: Cross-multiply to solve for \(h\).
Cross-multiplying gives:
\[ GM(R + h) = 2R \cdot GM \]
Cancelling \(GM\) (assuming \(GM \neq 0\)):
\[ R + h = 2R \]
Thus:
\[ h = 2R - R = R \]

### Conclusion:
The maximum height ascended by the particle is:
\[ \boxed{R} \]