Jordi stands 20 m from a wall and Diego stands 10 m from the same wall. Jordi throws a ball at an angle of 30 above the horizontal, and it collides elastically with the wall. How fast does Jordi need to throw the ball so that Diego will catch it? Consider Jordi and Diego to be the same height, and both are on the same perpendicular line from the wall.

$\textbf{(A) } 11 \text{ m/s}\\
\textbf{(B) }  15 \text{ m/s}\\
\textbf{(C) } 19 \text{ m/s}\\
\textbf{(D) } 30 \text{ m/s}\\
\textbf{(E) } 35 \text{ m/s}$
1. **Reflecting the Problem:**
   - When the ball collides elastically with the wall, it reflects off the wall. This reflection can be visualized as if Jordi is throwing the ball to a point 30 meters away (20 m to the wall + 10 m from the wall to Diego).

2. **Setting Up the Problem:**
   - Jordi throws the ball at an angle of \(30^\circ\) above the horizontal.
   - We need to find the initial velocity \(v\) such that the ball travels 30 meters horizontally and is caught by Diego.

3. **Horizontal and Vertical Components of Velocity:**
   - The horizontal component of the velocity is \(v \cos 30^\circ\).
   - The vertical component of the velocity is \(v \sin 30^\circ\).

4. **Time of Flight:**
   - The time of flight \(t\) can be found using the vertical motion equation. Since Jordi and Diego are at the same height, the ball will return to the same vertical level it was thrown from.
   - The time to reach the maximum height and come back down is given by:
     \[
     t = \frac{2 v \sin 30^\circ}{g}
     \]
     where \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)).

5. **Horizontal Distance:**
   - The horizontal distance covered by the ball is given by:
     \[
     d = v \cos 30^\circ \cdot t
     \]
   - Substituting the time of flight \(t\) into the horizontal distance equation:
     \[
     d = v \cos 30^\circ \cdot \frac{2 v \sin 30^\circ}{g}
     \]

6. **Simplifying the Equation:**
   - Using the trigonometric values \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \frac{1}{2}\):
     \[
     d = v \cdot \frac{\sqrt{3}}{2} \cdot \frac{2 v \cdot \frac{1}{2}}{g}
     \]
     \[
     d = v \cdot \frac{\sqrt{3}}{2} \cdot \frac{v}{g}
     \]
     \[
     d = \frac{v^2 \sqrt{3}}{2g}
     \]

7. **Solving for \(v\):**
   - We know the horizontal distance \(d\) is 30 meters:
     \[
     30 = \frac{v^2 \sqrt{3}}{2 \cdot 9.8}
     \]
     \[
     30 = \frac{v^2 \sqrt{3}}{19.6}
     \]
     \[
     v^2 \sqrt{3} = 30 \cdot 19.6
     \]
     \[
     v^2 \sqrt{3} = 588
     \]
     \[
     v^2 = \frac{588}{\sqrt{3}}
     \]
     \[
     v^2 = \frac{588 \cdot \sqrt{3}}{3}
     \]
     \[
     v^2 = 196 \sqrt{3}
     \]
     \[
     v = \sqrt{196 \sqrt{3}}
     \]
     \[
     v = 14 \sqrt[4]{3}
     \]
     \[
     v \approx 19 \, \text{m/s}
     \]

The final answer is \( \boxed{19 \, \text{m/s}} \)