We are analyzing the quadratic function $h(t) = -5t^2 + 20t + 3$, which represents the height of the ball after $t$ seconds. Let's solve each part step by step.

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a) Find the initial value when $t = 0$:

The initial value corresponds to the height of the ball when it is thrown ($t = 0$).

Substitute $t = 0$ into the function:

$h(0) = -5(0)^2 + 20(0) + 3 = 3$

Initial value: $h(0) = 3 \, \text{meters}$.

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b) Find the maximum/minimum value and when it occurs:

This is a quadratic function of the form $h(t) = at^2 + bt + c$, where $a = -5$, $b = 20$, and $c = 3$. Since $a < 0$, the parabola opens downward, meaning it has a maximum value.

The time at which the maximum occurs can be found using the formula for the vertex of a parabola:

$t_{\text{max}} = -\frac{b}{2a}$

Substitute $a = -5$ and $b = 20$:

$t_{\text{max}} = -\frac{20}{2(-5)} = 2 \, \text{seconds}$

Now, substitute $t = 2$ into the function to find the maximum height:

$h(2) = -5(2)^2 + 20(2) + 3$ $h(2) = -5(4) + 40 + 3 = -20 + 40 + 3 = 23$

Maximum height: $23 \, \text{meters}$, which occurs at $t = 2 \, \text{seconds}$.

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c) Given a $y$-value, find all possible $t$-values:

To find the times when the ball reaches a given height $h(t) = y$, solve the equation:

$-5t^2 + 20t + 3 = y$

Rearrange to standard quadratic form:

$-5t^2 + 20t + (3 - y) = 0$

This is a quadratic equation in $t$. Use the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = -5$, $b = 20$, and $c = 3 - y$. Substituting:

$t = \frac{-20 \pm \sqrt{20^2 - 4(-5)(3 - y)}}{2(-5)}$ $t = \frac{-20 \pm \sqrt{400 + 20(3 - y)}}{-10}$ $t = \frac{-20 \pm \sqrt{400 + 60 - 20y}}{-10}$ $t = \frac{-20 \pm \sqrt{460 - 20y}}{-10}$

Simplify further:

$t = \frac{20 \pm \sqrt{460 - 20y}}{10}$

The two solutions represent the two times the ball reaches the height $y$.

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Summary:

• (a): Initial height: $3 \, \text{meters}$.
• (b): Maximum height: $23 \, \text{meters}$, occurring at $t = 2 \, \text{seconds}$.
• (c): The times $t$ for a given height $y$ are:

$t = \frac{20 \pm \sqrt{460 - 20y}}{10}$

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Would you like a detailed example for part (c), or have any questions?

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5 Relative Questions:

1. How do you interpret the significance of the vertex in a quadratic motion problem?
2. Can the ball ever reach a height greater than $23 \, \text{meters}$? Why or why not?
3. How would the graph of $h(t)$ look, and what does it represent in real life?
4. What happens to the quadratic equation in part (c) if $y > 23$?
5. How would changing the coefficient $-5$ affect the trajectory of the ball?

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Tip:

In motion problems, the vertex of a parabola always represents the maximum or minimum point of the object's height, making it a key feature for analyzing such scenarios.