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 step by step, we will use the principles of conservation of energy and the formula for escape velocity.

### Step 1: Understand the given information
- The radius of the moon is \( R \).
- The depth of the crater is \( \frac{R}{100} \).
- The projectile is fired with a velocity equal to the escape velocity from the surface of the moon.

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

### Step 3: Apply conservation of energy
When the projectile is fired, it has kinetic energy and potential energy. As it rises, the kinetic energy will convert into potential energy until it reaches the maximum height where the velocity becomes zero.

The initial kinetic energy \( KE \) when the projectile is fired is:
\[ KE = \frac{1}{2} mv^2 \]
At the maximum height \( h \), the potential energy \( PE \) is given by:
\[ PE = -\frac{GMm}{R + h} \]
and at the depth of the crater, the potential energy is:
\[ PE_{\text{crater}} = -\frac{GMm}{R - \frac{R}{100}} = -\frac{GMm}{\frac{99R}{100}} = -\frac{100GMm}{99R} \]

### Step 4: Set up the energy conservation equation
Using the conservation of energy:
\[ KE_{\text{initial}} + PE_{\text{crater}} = PE_{\text{max height}} \]
Substituting the values we have:
\[ \frac{1}{2} mv^2 - \frac{100GMm}{99R} = -\frac{GMm}{R + h} \]

### Step 5: Substitute the escape velocity into the equation
Substituting \( v = \sqrt{\frac{2GM}{R}} \):
\[ \frac{1}{2} m \left(\frac{2GM}{R}\right) - \frac{100GMm}{99R} = -\frac{GMm}{R + h} \]
This simplifies to:
\[ \frac{GMm}{R} - \frac{100GMm}{99R} = -\frac{GMm}{R + h} \]

### Step 6: Simplify the equation
Combining the left side:
\[ \frac{GMm}{R} \left(1 - \frac{100}{99}\right) = -\frac{GMm}{R + h} \]
This gives:
\[ \frac{GMm}{R} \left(\frac{99 - 100}{99}\right) = -\frac{GMm}{R + h} \]
\[ -\frac{GMm}{99R} = -\frac{GMm}{R + h} \]

### Step 7: Cross-multiply and solve for \( h \)
Cross-multiplying gives:
\[ GMm(R + h) = 99GMm \]
Dividing by \( GMm \) (assuming \( m \neq 0 \)):
\[ R + h = 99R \]
Thus:
\[ h = 99R - R = 98R \]

### Step 8: Final height calculation
The maximum height attained by the projectile is:
\[ h = 98R \]

### Final Answer
The maximum height attained by the projectile is \( 98R \).