To determine how long after the ball is thrown it takes to reach a height of 7272 feet, we can set the given height equation equal to 7272 feet and solve for \( t \):
The height equation is:
\[ S(t) = 1212 + 6464t - 16t^2 \]
We need to set this equal to 7272:
\[ 1212 + 6464t - 16t^2 = 7272 \]
First, we will simplify this equation by moving all terms to one side:
\[ 1212 + 6464t - 16t^2 - 7272 = 0 \]
Combining like terms, we get:
\[ -16t^2 + 6464t - 6060 = 0 \]
To make the equation easier to work with, we'll multiply the equation by -1:
\[ 16t^2 - 6464t + 6060 = 0 \]
This is a quadratic equation in standard form \( at^2 + bt + c = 0 \) where:
- \( a = 16 \)
- \( b = -6464 \)
- \( c = 6060 \)
Next, we can use the quadratic formula to solve for \( t \):
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Calculating the discriminant:
\[ b^2 - 4ac = (-6464)^2 - 4(16)(6060) \]
Calculating \( (-6464)^2 \):
\[ 6464^2 = 41704576 \]
Calculating \( 4 \times 16 \times 6060 \):
\[ 4 \times 16 \times 6060 = 387840 \]
Now, subtract \( 387840 \) from \( 41704576 \):
\[ 41704576 - 387840 = 41316736 \]
Now, substitute back into the quadratic formula:
\[ t = \frac{6464 \pm \sqrt{41316736}}{32} \]
Calculating \( \sqrt{41316736} \):
\[ \sqrt{41316736} \approx 6426.43 \]
Now substituting this value back:
\[ t = \frac{6464 \pm 6426.43}{32} \]
Calculating the two possible values for \( t \):
-
For the positive root: \[ t = \frac{6464 + 6426.43}{32} \approx \frac{12890.43}{32} \approx 402.83 \]
-
For the negative root: \[ t = \frac{6464 - 6426.43}{32} \approx \frac{37.57}{32} \approx 1.17 \]
Thus, the ball reaches a height of 7272 feet at approximately 1.17 seconds and again at 402.83 seconds after being thrown.
So, your final answer is:
\[ \text{It takes approximately } 1.17 \text{ seconds and } 402.83 \text{ seconds for the ball to reach 7272 feet.} \]
In summary, the time it takes for the ball to reach the height of 7272 feet is approximately: \[ 1.17, 402.83 \]
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