Question
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.
make this like a 7th grader wrote it
\[
0 = -16t^2 + 20t + 3.
\]
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.
make this like a 7th grader wrote it
Answers
GPT-4o mini
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 \(\sqrt{37} \approx 6.08\)
1. For \( t_1 \):
\[
t_1 \approx \frac{5 + 6.08}{8} = \frac{11.08}{8} \approx 1.39.
\]
2. 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!
\[
0 = -16t^2 + 20t + 3.
\]
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 \(\sqrt{37} \approx 6.08\)
1. For \( t_1 \):
\[
t_1 \approx \frac{5 + 6.08}{8} = \frac{11.08}{8} \approx 1.39.
\]
2. 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!