5) A projectile is launched with speed $v_0$ off the edge of a cliff of height $h$, at an angle $\theta$ above the horizontal. Air friction is negligible. To maximize the horizontal range of the projectile, $\theta$ should satisfy which of the following?

A) $45^{\circ} < \theta < 90^{\circ}$
B) $\theta = 45^{\circ}$
C) $0^{\circ} < \theta < 45^{\circ}$
D) $\theta = 0^{\circ}$
E) $0^{\circ} < \theta < 45^{\circ}$ or $45^{\circ} < \theta < 90^{\circ}$ depending on the values of $h$ and $v_0$.
To solve this problem, we need to analyze the horizontal range of a projectile launched from a height \( h \) with an initial speed \( v_0 \) at an angle \( \theta \) above the horizontal. The goal is to determine the angle \( \theta \) that maximizes the horizontal range.

1. **Determine the time of flight:**
   The vertical motion of the projectile is governed by the equation:
   \[
   y(t) = v_0 \sin \theta \cdot t - \frac{1}{2} g t^2
   \]
   where \( y(t) \) is the vertical position at time \( t \), \( g \) is the acceleration due to gravity, and \( t \) is the time of flight. The projectile hits the ground when \( y(t) = -h \):
   \[
   -h = v_0 \sin \theta \cdot t - \frac{1}{2} g t^2
   \]
   Rearranging this quadratic equation, we get:
   \[
   \frac{1}{2} g t^2 - v_0 \sin \theta \cdot t - h = 0
   \]
   Solving for \( t \) using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = \frac{1}{2} g \), \( b = -v_0 \sin \theta \), and \( c = -h \):
   \[
   t = \frac{v_0 \sin \theta \pm \sqrt{(v_0 \sin \theta)^2 + 2gh}}{g}
   \]
   We take the positive root since time cannot be negative:
   \[
   t = \frac{v_0 \sin \theta + \sqrt{(v_0 \sin \theta)^2 + 2gh}}{g}
   \]

2. **Determine the horizontal range:**
   The horizontal range \( R \) is given by the horizontal velocity multiplied by the time of flight:
   \[
   R = v_0 \cos \theta \cdot t
   \]
   Substituting the expression for \( t \):
   \[
   R = v_0 \cos \theta \cdot \frac{v_0 \sin \theta + \sqrt{(v_0 \sin \theta)^2 + 2gh}}{g}
   \]
   Simplifying, we get:
   \[
   R = \frac{v_0^2 \sin \theta \cos \theta}{g} + \frac{v_0 \cos \theta \sqrt{(v_0 \sin \theta)^2 + 2gh}}{g}
   \]

3. **Analyze the range for different values of \( h \):**
   - When \( h \approx 0 \), the range equation simplifies to the standard projectile range on level ground:
     \[
     R = \frac{v_0^2 \sin 2\theta}{g}
     \]
     The range is maximized when \( \theta = 45^\circ \).

   - When \( h \) is very large, the term \( \sqrt{(v_0 \sin \theta)^2 + 2gh} \) dominates. In this case, the horizontal range is maximized by maximizing the horizontal component of the initial velocity, which occurs when \( \theta \) is close to \( 0^\circ \).

4. **Conclusion:**
   By continuity, for intermediate values of \( h \), the optimal angle \( \theta \) lies between \( 0^\circ \) and \( 45^\circ \). Therefore, the angle \( \theta \) that maximizes the horizontal range is in the interval \( 0^\circ < \theta < 45^\circ \).

The final answer is \( \boxed{\textbf{C}} \).