To find out when the soccer ball will hit the ground, we need to figure out when its height \( y \) is 0. We start with the equation:
\[ 0 = -16t^2 + 20t + 3. \]
This is a quadratic equation! To make it easier to work with, let's rearrange it to:
\[ -16t^2 + 20t + 3 = 0. \]
We can multiply everything by -1 to get rid of the negative sign:
\[ 16t^2 - 20t - 3 = 0. \]
Now, we can use the quadratic formula to find \( t \). The quadratic formula is:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
where \( a = 16 \), \( b = -20 \), and \( c = -3 \).
Step 1: Calculate the discriminant
First, let's find \( b^2 \):
\[ b^2 = (-20)^2 = 400. \]
Next, we need to calculate \( 4ac \):
\[ 4ac = 4 \cdot 16 \cdot (-3) = -192. \]
Now we can find out what's called the discriminant:
\[ b^2 - 4ac = 400 - (-192) = 400 + 192 = 592. \]
Step 2: Use the quadratic formula
Now we put our numbers into the quadratic formula:
\[ t = \frac{-(-20) \pm \sqrt{592}}{2 \cdot 16}. \]
This simplifies to:
\[ t = \frac{20 \pm \sqrt{592}}{32}. \]
Step 3: Simplify \(\sqrt{592}\)
Next, we’ll 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 have:
\[ t = \frac{20 \pm 4\sqrt{37}}{32}. \]
We can simplify this a bit more:
\[ t = \frac{20}{32} \pm \frac{4\sqrt{37}}{32} = \frac{5}{8} \pm \frac{\sqrt{37}}{8}. \]
Step 5: Calculate the two potential solutions
So, we have two different times:
\[ t_1 = \frac{5 + \sqrt{37}}{8}, \] \[ t_2 = \frac{5 - \sqrt{37}}{8}. \]
Step 6: Find the positive solution
Since time can’t be negative, let’s check which one is positive:
Estimating \(\sqrt{37} \approx 6.08\)
- For \( t_1 \):
\[ t_1 \approx \frac{5 + 6.08}{8} = \frac{11.08}{8} \approx 1.39. \]
- For \( t_2 \):
\[ t_2 \approx \frac{5 - 6.08}{8} = \frac{-1.08}{8} \approx -0.135. \]
Since time can’t be negative, we ignore \( t_2 \).
Conclusion
The soccer ball will hit the ground in about 1.39 seconds after being kicked!