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.

Determine the spring constant $k$.

$\textbf{(A) } \frac{mg}{h}\\ \\
\textbf{(B) }  \frac{2mg}{h}\\ \\
\textbf{(C) } \frac{mg}{H}\\ \\
\textbf{(D) } \frac{4mg}{H}\\ \\
\textbf{(E) } \frac{8mg}{H}$
1. **Identify the forces and energy involved:**
   - The man falls a distance \( H \) and comes to a momentary rest.
   - The bungee cord stretches from its original length \( L_0 \) to \( L = L_0 + h \).
   - 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 \( k \) is the spring constant and \( x \) is the extension.
   - At maximum extension \( h \), the tension in the bungee cord is \( T_{\text{max}} = kh \).

3. **Relate the maximum tension to the weight of the man:**
   \[
   kh = 4mg
   \]

4. **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 \( \frac{1}{2}kh^2 \).

5. **Set up the energy conservation equation:**
   \[
   mgH = \frac{1}{2}kh^2
   \]

6. **Substitute \( kh = 4mg \) into the energy equation:**
   \[
   mgH = \frac{1}{2} \cdot 4mg \cdot h
   \]
   \[
   mgH = 2mgh
   \]

7. **Solve for \( H \) in terms of \( h \):**
   \[
   H = 2h
   \]

8. **Express \( k \) in terms of \( mg \) and \( h \):**
   \[
   kh = 4mg \implies k = \frac{4mg}{h}
   \]

9. **Substitute \( h = \frac{H}{2} \) into the expression for \( k \):**
   \[
   k = \frac{4mg}{h} = \frac{4mg}{\frac{H}{2}} = \frac{4mg \cdot 2}{H} = \frac{8mg}{H}
   \]

Conclusion:
\[
k = \boxed{\frac{8mg}{H}}
\]