Solving Using Square Roots HC)

A ball drops from the top of a building 96 feet above the ground. The function h(t) = –16t 2 + 96 models the height, h(t), in feet, of the ball after t seconds.

Part A: For what values of t is the ball at ground level? Show all work. (3 points)

Part B: Explain which solutions to h(t) are viable and nonviable. (1 point)



We are given the height function \( h(t) = -16t^2 + 96 \), which models the height of a ball dropped from 96 feet. The task is to determine when the ball hits the ground (Part A) and to discuss the viability of the solutions (Part B).

### Part A: For what values of \( t \) is the ball at ground level?

The ball reaches the ground when its height is 0, meaning we set \( h(t) = 0 \) and solve for \( t \).

\[
h(t) = -16t^2 + 96
\]
Set \( h(t) = 0 \):

\[
0 = -16t^2 + 96
\]

1. Move the constant to the other side:
\[
16t^2 = 96
\]

2. Divide by 16:
\[
t^2 = \frac{96}{16} = 6
\]

3. Solve for \( t \) by taking the square root of both sides:
\[
t = \pm\sqrt{6}
\]
\[
t = \pm 2.45 \quad (\text{approximately})
\]

So, the possible values of \( t \) are \( t = 2.45 \) seconds and \( t = -2.45 \) seconds.

### Part B: Explain which solutions to \( h(t) \) are viable and nonviable.

- The negative value \( t = -2.45 \) seconds is **nonviable** because time cannot be negative in this context (we're measuring time after the ball is dropped).
- The positive value \( t = 2.45 \) seconds is **viable**, meaning the ball hits the ground approximately 2.45 seconds after it is dropped.

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Would you like more details on the reasoning behind this solution? Here are some related questions to consider:

1. How would the function change if the ball was dropped from a different height?
2. What would happen to the equation if air resistance was considered?
3. How can we interpret the constant \(-16\) in the function \( h(t) = -16t^2 + 96 \)?
4. What does the graph of \( h(t) = -16t^2 + 96 \) look like, and how does it help understand the motion of the ball?
5. How can you calculate the maximum height the ball reaches if thrown upwards instead of dropped?

**Tip**: In problems involving gravity, the acceleration constant \(-16\) (in feet per second squared) represents half of Earth's gravitational acceleration, which simplifies the equation for height

A ball drops from the top of a building 96 feet above the ground. The function h(t) = –16t^2 + 96 models the height, h(t), in feet, of the ball after t seconds.

Part A: For what values of t is the ball at ground level? Show all work.

Part B: Explain which solutions to h(t) are viable and nonviable.

Learn how to determine when a ball dropped from a 96-foot building reaches the ground using the equation h(t) = -16t^2 + 96. This problem covers solving quadratic equations using square roots, and discusses viable solutions based on physical constraints. Suitable for students in Grades 8-10.

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