Let's consider the scenario of a soccer ball being kicked upwards into the air from a height of 3 feet. We can model the height of the soccer ball above the ground as a quadratic function using the formula \( y = -16t^2 + vt + h \), where \( v \) is the initial velocity (in feet per second) when the ball is kicked, and \( h \) is the initial height (in feet). For this example, let's assume the initial velocity of the ball is 20 feet per second. Therefore, the complete function becomes:

Question

\[ y = -16t^2 + 20t + 3. \]

In this function, \( y \) represents the height of the soccer ball in feet, and \( t \) is the time in seconds after the ball has been kicked. The constant value of -16 represents the acceleration due to gravity in feet per second squared (the negative sign indicates that the height decreases over time), while the initial height \( h \) of 3 feet indicates the starting position from which the ball is kicked. To predict when the ball will hit the ground, we would set \( y = 0 \) and solve for \( t \).

Answer 1

To determine when the soccer ball will hit the ground, we need to set the height \( y \) equal to 0 and solve for \( t \) in the quadratic equation:

\[ 0 = -16t^2 + 20t + 3. \]

This can be rearranged to the standard form of a quadratic equation:

\[ -16t^2 + 20t + 3 = 0. \]

To make it easier to work with, we can multiply the entire equation by -1 to eliminate the negative leading coefficient:

\[ 16t^2 - 20t - 3 = 0. \]

Now, we will use the quadratic formula, which is given by:

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

where \( a = 16 \), \( b = -20 \), and \( c = -3 \).

Step 1: Calculate the discriminant (\( b^2 - 4ac \))

First, compute \( b^2 \):

\[ b^2 = (-20)^2 = 400. \]

Next, compute \( 4ac \):

\[ 4ac = 4 \cdot 16 \cdot (-3) = -192. \]

Now, calculate the discriminant:

\[ b^2 - 4ac = 400 - (-192) = 400 + 192 = 592. \]

Step 2: Apply the quadratic formula

Now substitute the values into the quadratic formula:

\[ t = \frac{-(-20) \pm \sqrt{592}}{2 \cdot 16}, \]

which simplifies to:

\[ t = \frac{20 \pm \sqrt{592}}{32}. \]

Step 3: Simplify \(\sqrt{592}\)

To simplify \(\sqrt{592}\):

\[ 592 = 16 \cdot 37, \] so:

\[ \sqrt{592} = \sqrt{16 \cdot 37} = 4\sqrt{37}. \]

Step 4: Substitute back into the formula

Now we can write:

\[ t = \frac{20 \pm 4\sqrt{37}}{32}. \]

This can be further simplified:

\[ t = \frac{20}{32} \pm \frac{4\sqrt{37}}{32} = \frac{5}{8} \pm \frac{\sqrt{37}}{8}. \]

Step 5: Calculate the two potential solutions

Thus, we have:

\[ t_1 = \frac{5 + \sqrt{37}}{8}, \] \[ t_2 = \frac{5 - \sqrt{37}}{8}. \]

Step 6: Determine the positive solution

Since we're looking for a positive time, we consider:

Estimate \(\sqrt{37} \approx 6.08\)

Calculating:

1. For \( t_1 \):

\[ t_1 \approx \frac{5 + 6.08}{8} = \frac{11.08}{8} \approx 1.385. \]

2. For \( t_2 \):

\[ t_2 \approx \frac{5 - 6.08}{8} = \frac{-1.08}{8} \approx -0.135. \]

Since time cannot be negative, we have:

\[ t \approx 1.385 \text{ seconds}. \]

Conclusion

The soccer ball will hit the ground approximately 1.39 seconds after being kicked.