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To determine when the rocket will hit the ground, we need to find the value of \( s \) that makes the height function \( h(s) = -3s^2 + 6s + 45 \) equal to zero (since the height will be zero when the rocket reaches the ground). 

Let's solve the equation:

\[
-3s^2 + 6s + 45 = 0
\]

### Step 1: Simplify the Equation
Divide both sides by \(-3\):

\[
s^2 - 2s - 15 = 0
\]

### Step 2: Factor the Quadratic Equation
Now, factor the quadratic:

\[
s^2 - 2s - 15 = (s - 5)(s + 3) = 0
\]

### Step 3: Solve for \( s \)
Set each factor equal to zero:

1. \( s - 5 = 0 \Rightarrow s = 5 \)
2. \( s + 3 = 0 \Rightarrow s = -3 \)

Since time cannot be negative, we discard \( s = -3 \). Therefore, the rocket will hit the ground after **5 seconds**.

### Answer:
The correct choice is **5 seconds**.

Would you like further details or explanations?

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Here are some related questions for practice:

1. How would the answer change if the coefficient of \( s^2 \) were different?
2. What if the initial height was lower? How would that affect the time?
3. How does changing the value of the linear term affect the height function?
4. What are the solutions to \( h(s) = 0 \) if the equation cannot be factored?
5. What if the function represented a projectile launched downward?

**Tip:** Always check if a quadratic equation can be factored easily before applying the quadratic formula.

For a science project, Joseph launches a model rocket. The function h(s) = -3s^2 + 6s + 45 represents the height in feet of the rocket after s seconds. It will take the rocket ___ second(s) to hit the ground.

Learn how to calculate the time it takes for a model rocket to hit the ground using the quadratic function h(s) = -3s^2 + 6s + 45. This problem involves factoring and applying the zero product property, suitable for high school students in Grades 9-11.

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