Bob, a spherical person, is floating around peacefully when Dave the giant orange fish launches him straight up 23 m/s with his tail. If Bob has density 100 $\text{kg/m}^3$, let $f(r)$ denote how far underwater his centre of mass plunges underwater once he lands, assuming his centre of mass was at water level when he's launched up. Find $\lim_{r\to0} \left(f(r)\right) $. Express your answer is meters and round to the nearest integer. Assume the density of water is 1000 $\text{kg/m}^3$. 

[i](B. Dejean, 6 points)[/i]
1. **Initial Setup and Assumptions:**
   - Bob is launched upwards with an initial velocity \( u = 23 \, \text{m/s} \).
   - Bob's density \( \sigma = 100 \, \text{kg/m}^3 \).
   - Water's density \( \rho = 1000 \, \text{kg/m}^3 \).
   - We need to find the limit of \( f(r) \) as \( r \to 0 \), where \( f(r) \) is the depth Bob's center of mass plunges underwater.

2. **Energy Conservation:**
   - Using the work-energy theorem, the initial kinetic energy of Bob is given by:
     \[
     \frac{1}{2} m u^2
     \]
   - The work done by the buoyant force \( W_{\text{buy}} \) and the gravitational potential energy change \( mgh \) must be considered.

3. **Buoyant Force and Work Done:**
   - The buoyant force \( F_{\text{buy}} \) is given by Archimedes' principle:
     \[
     F_{\text{buy}} = \rho V g
     \]
     where \( V \) is the volume of Bob.
   - The work done by the buoyant force as Bob plunges to depth \( h \) is:
     \[
     W_{\text{buy}} = F_{\text{buy}} \cdot h = \rho V g h
     \]

4. **Volume and Mass Relationship:**
   - Bob's volume \( V \) can be expressed in terms of his mass \( m \) and density \( \sigma \):
     \[
     V = \frac{m}{\sigma}
     \]

5. **Energy Conservation Equation:**
   - Equating the initial kinetic energy to the work done by the buoyant force and the gravitational potential energy change:
     \[
     \frac{1}{2} m u^2 = \rho V g h - mgh
     \]
   - Substituting \( V = \frac{m}{\sigma} \):
     \[
     \frac{1}{2} m u^2 = \rho \left( \frac{m}{\sigma} \right) g h - mgh
     \]
   - Simplifying:
     \[
     \frac{1}{2} u^2 = \left( \frac{\rho}{\sigma} - 1 \right) gh
     \]

6. **Solving for \( h \):**
   - Rearranging the equation to solve for \( h \):
     \[
     h = \frac{\frac{1}{2} u^2}{\left( \frac{\rho}{\sigma} - 1 \right) g}
     \]
   - Substituting the given values \( u = 23 \, \text{m/s} \), \( \rho = 1000 \, \text{kg/m}^3 \), \( \sigma = 100 \, \text{kg/m}^3 \), and \( g = 9.81 \, \text{m/s}^2 \):
     \[
     h = \frac{\frac{1}{2} \cdot 23^2}{\left( \frac{1000}{100} - 1 \right) \cdot 9.81}
     \]
     \[
     h = \frac{\frac{1}{2} \cdot 529}{(10 - 1) \cdot 9.81}
     \]
     \[
     h = \frac{264.5}{9 \cdot 9.81}
     \]
     \[
     h = \frac{264.5}{88.29}
     \]
     \[
     h \approx 2.99 \, \text{m}
     \]

The final answer is \( \boxed{3} \, \text{m} \).