A trapezoid has area $2\, m^2$ and the sum of its diagonals is $4\,m$. Determine the height of this trapezoid.
1. **Given Information:**
   - The area of the trapezoid is \(2 \, \text{m}^2\).
   - The sum of the lengths of the diagonals is \(4 \, \text{m}\).

2. **Using the formula for the area of a trapezoid in terms of its diagonals:**
   - Let \(l_1\) and \(l_2\) be the lengths of the diagonals.
   - Let \(\alpha\) be the angle between the diagonals.
   - The area \(A\) of the trapezoid can be given by:
     \[
     A = \frac{1}{2} l_1 l_2 \sin \alpha
     \]
   - Given \(A = 2 \, \text{m}^2\), we have:
     \[
     2 = \frac{1}{2} l_1 l_2 \sin \alpha
     \]
     \[
     4 = l_1 l_2 \sin \alpha
     \]

3. **Using the sum of the diagonals:**
   - Given \(l_1 + l_2 = 4 \, \text{m}\).

4. **Maximizing the product \(l_1 l_2\):**
   - By the Arithmetic Mean-Geometric Mean Inequality (AM-GM Inequality):
     \[
     \frac{l_1 + l_2}{2} \geq \sqrt{l_1 l_2}
     \]
     \[
     \frac{4}{2} \geq \sqrt{l_1 l_2}
     \]
     \[
     2 \geq \sqrt{l_1 l_2}
     \]
     \[
     4 \geq l_1 l_2
     \]
   - Since \(4 = l_1 l_2 \sin \alpha\), the maximum value of \(l_1 l_2\) is \(4\) when \(\sin \alpha = 1\), which implies \(\alpha = 90^\circ\).

5. **Finding the lengths of the diagonals:**
   - If \(\alpha = 90^\circ\), then:
     \[
     4 = l_1 l_2
     \]
   - Given \(l_1 + l_2 = 4\), we solve the system of equations:
     \[
     l_1 + l_2 = 4
     \]
     \[
     l_1 l_2 = 4
     \]
   - Solving these, we find \(l_1 = l_2 = 2 \, \text{m}\).

6. **Finding the sum of the bases:**
   - The sum of the bases \(b_1 + b_2\) can be found using the relationship:
     \[
     b_1 + b_2 = \sqrt{l_1^2 + l_2^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \, \text{m}
     \]

7. **Finding the height \(h\) of the trapezoid:**
   - The height \(h\) can be found using the area formula for a trapezoid:
     \[
     A = \frac{1}{2} (b_1 + b_2) h
     \]
     \[
     2 = \frac{1}{2} (2\sqrt{2}) h
     \]
     \[
     2 = \sqrt{2} h
     \]
     \[
     h = \frac{2}{\sqrt{2}} = \sqrt{2} \, \text{m}
     \]

The final answer is \( \boxed{ \sqrt{2} \, \text{m} } \)