Joe throws a ball upwards from a certain height above the ground level. The height of the ball above the ground after time t seconds from when the ball was thrown is given by the expression `h(t)= -(t-a)^(2)+b`. The ball reaches a maximum height of 25 feet after 4 seconds. After how much time [in seconds] will the ball reach the ground level?

Options

A. 6
B. 7
C. 8
D. 9

Answer

To solve the problem, we need to analyze the given height function of the ball, which is represented as:

\[ h(t) = -(t - a)^2 + b \]

### Step 1: Identify the maximum height and time
We know that the ball reaches a maximum height of 25 feet after 4 seconds. This means:

- The maximum height \( h(t) \) occurs at \( t = 4 \).
- At this time, \( h(4) = 25 \).

### Step 2: Determine the value of \( a \)
Since the maximum height occurs at \( t = a \), we can conclude that:

\[ a = 4 \]

### Step 3: Substitute \( a \) into the height function
Now we substitute \( a \) back into the height function:

\[ h(t) = -(t - 4)^2 + b \]

### Step 4: Find the value of \( b \)
We know that at \( t = 4 \), the height \( h(4) = 25 \):

\[ h(4) = -(4 - 4)^2 + b = 25 \]

This simplifies to:

\[ 0 + b = 25 \]

Thus, we find:

\[ b = 25 \]

### Step 5: Write the complete height function
Now we can write the complete height function:

\[ h(t) = -(t - 4)^2 + 25 \]

### Step 6: Determine when the ball reaches the ground
The ball reaches the ground when \( h(t) = 0 \):

\[ 0 = -(t - 4)^2 + 25 \]

### Step 7: Rearranging the equation
Rearranging gives us:

\[ (t - 4)^2 = 25 \]

### Step 8: Solve for \( t \)
Taking the square root of both sides, we get:

\[ t - 4 = 5 \quad \text{or} \quad t - 4 = -5 \]

This leads to two possible solutions:

1. \( t - 4 = 5 \)
\[ t = 9 \]

2. \( t - 4 = -5 \)
\[ t = -1 \] (This solution is not valid since time cannot be negative.)

### Conclusion
Thus, the time when the ball reaches the ground level is:

\[ t = 9 \text{ seconds} \]