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