Amanita, let's carefully solve the problem inside the green rectangle, which is problem 13. Here’s what the problem says and how we’ll solve it step by step:

Problem Summary: 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:

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

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

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Step-by-Step Solution Plan

Step 1: Understand the formula

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

Step 2: Find the time when the ball reaches its maximum height

• The maximum value of a quadratic $x = at^2 + bt + c$ (with $a < 0$) occurs at $t = -\frac{b}{2a}$.
• Here, $a = -5$, $b = 20$.

Step 3: Calculate the time $t$ at which the maximum height occurs

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

Step 4: Find the maximum height

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

Step 5: Check if the time is within the allowed range ($0 \leq t \leq 4$)

• If it is, that’s our answer. If not, check the height at the endpoints $t = 0$ and $t = 4$.

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Now, let's execute each step in detail.

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Step 3: Calculate the time $t$ at which the maximum height occurs

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

Let's calculate this value. Amanita, let's continue with the solution:

From our calculation, the time when the ball reaches its maximum height is:

$$t = 2 \text{ seconds}$$

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

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

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

$$x = 40 - 20 = 20$$

So, the maximum height the ball reaches is:

$$\boxed{20 \text{ meters}}$$

This value is within the allowed time range ($0 \leq t \leq 4$), so it is our final answer.