A man with mass $m$ jumps off of a high bridge with a bungee cord attached to his ankles. The man falls through a maximum distance $H$ at which point the bungee cord brings him to a momentary rest before he bounces back up. The bungee cord is perfectly elastic, obeying Hooke's force law with a spring constant $k$, and stretches from an original length of $L_0$ to a final length $L = L_0 + h$. The maximum tension in the Bungee cord is four times the weight of the man.

Find the maximum extension of the bungee cord $h$.

$\textbf{(A) } h = \frac{1}{2}H \\ \\ 
\textbf{(B) }  h = \frac{1}{4}H\\ \\
\textbf{(C) } h = \frac{1}{5}H\\ \\
\textbf{(D) } h = \frac{2}{5}H\\ \\
\textbf{(E) } h = \frac{1}{8}H$
1. **Identify the forces and energy involved:**
   - The man has a mass \( m \) and falls a maximum distance \( H \).
   - The bungee cord has an original length \( L_0 \) and stretches to a final length \( L = L_0 + h \).
   - The bungee cord obeys Hooke's law with a spring constant \( k \).
   - The maximum tension in the bungee cord is four times the weight of the man, i.e., \( T_{\text{max}} = 4mg \).

2. **Apply Hooke's Law:**
   - Hooke's law states that the force exerted by a spring is \( F = kx \), where \( x \) is the extension of the spring.
   - At maximum extension \( h \), the tension in the bungee cord is \( T_{\text{max}} = kh \).
   - Given \( T_{\text{max}} = 4mg \), we have:
     \[
     kh = 4mg
     \]
     \[
     h = \frac{4mg}{k}
     \]

3. **Use conservation of energy:**
   - The potential energy lost by the man falling a distance \( H \) is converted into the elastic potential energy stored in the bungee cord.
   - The gravitational potential energy lost is \( mgh \).
   - The elastic potential energy stored in the bungee cord at maximum extension is \( \frac{1}{2}kh^2 \).
   - By conservation of energy:
     \[
     mgH = \frac{1}{2}kh^2
     \]

4. **Substitute \( h \) from Hooke's law into the energy equation:**
   - From \( kh = 4mg \), we have \( k = \frac{4mg}{h} \).
   - Substitute \( k \) into the energy equation:
     \[
     mgH = \frac{1}{2} \left( \frac{4mg}{h} \right) h^2
     \]
     \[
     mgH = \frac{1}{2} \cdot 4mg \cdot h
     \]
     \[
     mgH = 2mg \cdot h
     \]
     \[
     H = 2h
     \]
     \[
     h = \frac{H}{2}
     \]

5. **Conclusion:**
   - The maximum extension of the bungee cord \( h \) is \( \frac{H}{2} \).

The final answer is \(\boxed{h = \frac{1}{2}H}\).