1. Given:

   • Initial height from which the ball is dropped: $h_1 = 20\,\text{m}$
   • Coefficient of restitution: $e = 0.5$

2. Relation between rebound speed and impact speed: For collision with the floor, $e = \frac{v_{\text{after}}}{v_{\text{before}}}$ so, $v_{\text{after}} = e\,v_{\text{before}}$

3. Speed just before hitting the floor: Using energy (or kinematics), $v_{\text{before}}^2 = 2gh_1 = 2g(20)$

4. Speed just after rebound: $v_{\text{after}} = e v_{\text{before}}$ Squaring, $v_{\text{after}}^2 = e^2 v_{\text{before}}^2 = e^2(2gh_1)$

5. Maximum height after rebound: Let rebound height be $h_2$. Then, $v_{\text{after}}^2 = 2gh_2$ Hence, $2gh_2 = e^2(2gh_1)$ $h_2 = e^2 h_1$

6. Substitute values: $h_2 = (0.5)^2 \times 20 = 0.25 \times 20 = 5\,\text{m}$

Therefore, the ball rebounds to a height of $\boxed{5\,\text{m}}$