When a ball is thrown straight up at an initial velocity of 54 feet per second. The height of the ball t seconds after it is thrown is given by the function `h(t)=54t-12t^(2)`. How many seconds after the ball is thrown will it return to the ground?

Options

A. `9.2`
B. `6`
C. `4.5`
D. `4`

Answer

To find out how many seconds after the ball is thrown it will return to the ground, we need to analyze the height function given by:

\[ h(t) = 54t - 12t^2 \]

### Step 1: Set the height function equal to zero
To determine when the ball returns to the ground, we need to find the time \( t \) when the height \( h(t) \) is zero. Therefore, we set the equation:

\[ 54t - 12t^2 = 0 \]

### Step 2: Factor the equation
We can factor out \( t \) from the equation:

\[ t(54 - 12t) = 0 \]

### Step 3: Solve for \( t \)
This gives us two solutions:

1. \( t = 0 \) (the time when the ball is thrown)
2. \( 54 - 12t = 0 \)

Now, solving the second equation:

\[ 54 = 12t \]

Dividing both sides by 12:

\[ t = \frac{54}{12} \]

### Step 4: Simplify the fraction
Now, simplify \( \frac{54}{12} \):

\[ t = \frac{27}{6} = \frac{9}{2} \]

### Step 5: Convert to decimal
Convert \( \frac{9}{2} \) to a decimal:

\[ t = 4.5 \]

### Conclusion
The ball will return to the ground after **4.5 seconds**.

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