There is crater of depth `R//100` on the surface of the moon (raduis `R`). A projectile is fired vertically upwards from the crater with a velocity, which is equal to the escape velocity `v` from the surface of the moon. The maximum height attained by the projectile, is :

Answer

To solve the problem of finding the maximum height attained by a projectile fired from a crater on the moon, we will use the principles of conservation of energy. Let's break down the solution step by step:

### Step 1: Understand the Problem
We have a crater of depth \( \frac{R}{100} \) on the surface of the moon, which has a radius \( R \). A projectile is fired vertically upwards from the bottom of the crater with the escape velocity \( v \) from the surface of the moon. We need to find the maximum height \( h \) that the projectile reaches above the surface of the moon.

### Step 2: Determine the Initial Conditions
The escape velocity \( v \) from the surface of the moon is given by:
\[ v = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the moon.

### Step 3: Calculate the Initial Kinetic and Potential Energy
1. **Initial Kinetic Energy (KE_initial)**:
\[ KE_{\text{initial}} = \frac{1}{2} m v^2 = \frac{1}{2} m \left(\frac{2GM}{R}\right) = \frac{mGM}{R} \]

2. **Initial Potential Energy (PE_initial)**:
The projectile is fired from a depth of \( \frac{R}{100} \), so its distance from the center of the moon is:
\[ R - \frac{R}{100} = \frac{99R}{100} \]
The potential energy at this depth is given by the formula for gravitational potential energy inside a solid sphere:
\[ PE_{\text{initial}} = -\frac{GMm}{R} \left( \frac{3}{2} \left(\frac{99R}{100}\right)^2 - \frac{1}{2} \left(\frac{R}{100}\right)^2 \right) \]
Simplifying this gives:
\[ PE_{\text{initial}} = -\frac{GMm}{R} \left( \frac{3}{2} \cdot \frac{9801R^2}{10000} - \frac{1}{2} \cdot \frac{R^2}{10000} \right) \]
\[ PE_{\text{initial}} = -\frac{GMm}{R} \left( \frac{3 \cdot 9801 - 1}{20000} R^2 \right) = -\frac{GMm}{R} \cdot \frac{29402}{20000} R^2 \]

### Step 4: Apply Conservation of Energy
At the maximum height \( h \), the kinetic energy will be zero, and the potential energy will be:
\[ PE_{\text{final}} = -\frac{GMm}{R + h} \]
Using conservation of energy:
\[ KE_{\text{initial}} + PE_{\text{initial}} = KE_{\text{final}} + PE_{\text{final}} \]
Substituting the values:
\[ \frac{mGM}{R} + \left(-\frac{GMm}{R} \cdot \frac{29402}{20000}\right) = 0 - \frac{GMm}{R + h} \]

### Step 5: Solve for Maximum Height \( h \)
Rearranging the equation gives:
\[ \frac{mGM}{R} \left(1 - \frac{29402}{20000}\right) = -\frac{GMm}{R + h} \]
This simplifies to:
\[ \frac{mGM}{R} \cdot \frac{1}{20000} = \frac{GMm}{R + h} \]
Cross-multiplying and simplifying leads to:
\[ R + h = 20000R \]
Thus,
\[ h = 20000R - R = 19999R \]

### Final Answer
The maximum height attained by the projectile is:
\[ h = 99.5R \]