Amanita, let's solve the problem inside the green rectangle step by step. The problem is about finding the maximum height reached by a ball thrown upward from the ground with a speed of 20 m/s. The height $x$ (in meters) after $t$ seconds is given by:

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

We are asked to find the maximum height the ball reaches, for $0 \leq t \leq 4$.

Step-by-Step Plan

1. Understand the formula: The formula $x = 20t - 5t^2$ describes the height of the ball at any time $t$. This is a quadratic equation, which forms a parabola opening downwards (because the $t^2$ term is negative).

2. Find the time when the ball reaches its maximum height: The maximum value of a quadratic equation $ax^2 + bx + c$ (when $a < 0$) occurs at $t = -\frac{b}{2a}$. Here, $a = -5$ and $b = 20$.

3. Calculate the maximum height: Substitute the value of $t$ found in step 2 back into the original equation to find the maximum height.

4. Check the interval: Make sure the time $t$ for the maximum height is within the allowed range $0 \leq t \leq 4$.

Let's start by finding the time when the ball reaches its maximum height. Amanita, let's go through the solution step by step:

1. The height of the ball after $t$ seconds is given by:

$x = 20t - 5t^2$

This is a quadratic equation, and its graph is a parabola that opens downward.

2. The maximum height is reached at the vertex of the parabola. The time $t$ at which the maximum occurs is:

$$t = -\frac{b}{2a}$$

For our equation, $a = -5$ and $b = 20$:

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

So, the ball reaches its maximum height at $t = 2$ seconds.

3. Substitute $t = 2$ into the original equation to find the maximum height:

$x = 20 \times 2 - 5 \times (2^2) = 40 - 20 = 20$

4. The time $t = 2$ is within the allowed range $0 \leq t \leq 4$.

Final Answer:
The highest point the ball reaches is $\boxed{20}$ meters.