`h(t)=144t-16t^(2)`
The function above represents the height, in feet, a ball reaches t seconds after it is tossed in the air from groud level.
a. What is the maximum height of the ball?
b. After how many seconds will the ball hit the ground before rebounding?

Answer

To solve the problem step by step, we will break it down into two parts: finding the maximum height of the ball and determining when the ball will hit the ground.

### Part a: Finding the Maximum Height

1. **Identify the function**: The height of the ball as a function of time \( t \) is given by:
\[ h(t) = 144t - 16t^2 \]

2. **Differentiate the function**: To find the maximum height, we need to differentiate \( h(t) \) with respect to \( t \):
\[ h'(t) = \frac{d}{dt}(144t - 16t^2) = 144 - 32t \]

3. **Set the derivative to zero**: To find the critical points, we set the derivative equal to zero:
\[ 144 - 32t = 0 \]

4. **Solve for \( t \)**: Rearranging the equation gives:
\[ 32t = 144 \quad \Rightarrow \quad t = \frac{144}{32} = \frac{9}{2} = 4.5 \text{ seconds} \]

5. **Substitute \( t \) back into the height function**: Now we substitute \( t = 4.5 \) back into the original height function to find the maximum height:
\[ h\left(\frac{9}{2}\right) = 144\left(\frac{9}{2}\right) - 16\left(\frac{9}{2}\right)^2 \]
\[ = 144 \times 4.5 - 16 \times 20.25 \]
\[ = 648 - 324 = 324 \text{ feet} \]

### Part b: Finding When the Ball Hits the Ground

1. **Set the height function to zero**: The ball hits the ground when the height is zero:
\[ h(t) = 0 \quad \Rightarrow \quad 144t - 16t^2 = 0 \]

2. **Factor the equation**: We can factor out \( t \):
\[ t(144 - 16t) = 0 \]

3. **Solve for \( t \)**: This gives us two solutions:
\[ t = 0 \quad \text{or} \quad 144 - 16t = 0 \]
Solving for \( t \) in the second equation:
\[ 16t = 144 \quad \Rightarrow \quad t = \frac{144}{16} = 9 \text{ seconds} \]

### Final Answers

- **Maximum height of the ball**: 324 feet
- **Time when the ball hits the ground**: 9 seconds