Amanita, let's carefully solve the problem step by step.

Understanding the Problem:

• A ball is thrown straight up from the ground with a speed of $20$ m/s.
• The height $x$ (in meters) after $t$ seconds is given by the formula:

$$x = 20t - 5t^2$$

• We are asked to find the maximum height the ball reaches.

Step-by-Step Plan:

1. Recognize the Type of Equation:

   • The formula $x = 20t - 5t^2$ is a quadratic equation in terms of $t$.
   • The graph of this equation is a parabola that opens downward (because the coefficient of $t^2$ is negative).

2. Find the Time When the Ball Reaches Maximum Height:

   • The maximum value of a quadratic equation $ax^2 + bx + c$ occurs at $t = -\frac{b}{2a}$.
   • Here, $a = -5$ and $b = 20$.

3. Calculate the Time for Maximum Height:

   • Substitute the values into the formula to find $t$.

4. Find the Maximum Height:

   • Substitute the value of $t$ back into the original equation to find the maximum height $x$.

Let's execute these steps.

Step 1: Find the time when the ball reaches its maximum height.

$t = -\frac{b}{2a} = -\frac{20}{2 \times (-5)}$

Step 2: Calculate the value of $t$.

Amanita, let's continue with the solution:

• From our calculation, the time when the ball reaches its maximum height is $t = 2$ seconds.

Now, let's find the maximum height by substituting $t = 2$ into the height formula:

$$x = 20t - 5t^2$$

$x = 20 \times 2 - 5 \times (2^2)$

$$x = 40 - 20 = 20$$

Final Answer:
The maximum height the ball reaches is $20$ meters.