A soccer ball is kicked upward from gound level with an initial velocity of 52 feet per second. The function `h(t)=-16t^2+52t` gives the ball's height , in feet , after t seconds. For how many seconds, to the nearest tenth of a second , is the ball at least 20 feet above the ground ?

Answer

To solve the problem, we need to determine the time intervals during which the soccer ball is at least 20 feet above the ground. We start with the height function given by:

\[ h(t) = -16t^2 + 52t \]

We want to find the values of \( t \) for which \( h(t) \geq 20 \). This leads us to set up the inequality:

\[ -16t^2 + 52t \geq 20 \]

### Step 1: Set up the equation
First, we rearrange the inequality to form a standard quadratic equation:

\[ -16t^2 + 52t - 20 \geq 0 \]

### Step 2: Rewrite in standard form
Next, we can express this as:

\[ 16t^2 - 52t + 20 \leq 0 \]

### Step 3: Divide by 4
To simplify the equation, we can divide all terms by 4:

\[ 4t^2 - 13t + 5 \leq 0 \]

### Step 4: Find the roots
Now we need to find the roots of the quadratic equation \( 4t^2 - 13t + 5 = 0 \) using the quadratic formula:

\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 4 \), \( b = -13 \), and \( c = 5 \).

### Step 5: Calculate the discriminant
Calculating the discriminant:

\[ b^2 - 4ac = (-13)^2 - 4 \cdot 4 \cdot 5 = 169 - 80 = 89 \]

### Step 6: Apply the quadratic formula
Now we can find the roots:

\[ t = \frac{13 \pm \sqrt{89}}{8} \]

### Step 7: Calculate the roots
Calculating the two possible values for \( t \):

1. \( t_1 = \frac{13 + \sqrt{89}}{8} \)
2. \( t_2 = \frac{13 - \sqrt{89}}{8} \)

Using a calculator, we find:

- \( \sqrt{89} \approx 9.434 \)
- \( t_1 \approx \frac{13 + 9.434}{8} \approx \frac{22.434}{8} \approx 2.80425 \)
- \( t_2 \approx \frac{13 - 9.434}{8} \approx \frac{3.566}{8} \approx 0.44575 \)

### Step 8: Determine the interval
The ball is above 20 feet between the two roots:

\[ 0.44575 \leq t \leq 2.80425 \]

### Step 9: Round to the nearest tenth
To find the duration for which the ball is at least 20 feet above the ground, we calculate the difference between the two times:

\[ 2.80425 - 0.44575 \approx 2.3585 \]

Rounding to the nearest tenth, we find that the ball is at least 20 feet above the ground for approximately:

**2.4 seconds.**