To find how long it takes for the ball to reach a height of 7272 feet, we can use the given equation for the height of the ball:
\[ S(t) = 1212 + 6464t - 16t^2 \]
We want to find \( t \) such that \( S(t) = 7272 \). Thus, we set up the equation:
\[ 7272 = 1212 + 6464t - 16t^2 \]
Subtract 1212 from both sides:
\[ 7272 - 1212 = 6464t - 16t^2 \]
This simplifies to:
\[ 6060 = 6464t - 16t^2 \]
Rearranging gives us:
\[ 16t^2 - 6464t + 6060 = 0 \]
Now, we can solve this quadratic equation using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 16 \), \( b = -6464 \), and \( c = 6060 \). Now we calculate the discriminant (\( b^2 - 4ac \)):
\[ b^2 - 4ac = (-6464)^2 - 4 \cdot 16 \cdot 6060 \]
Calculating \( (-6464)^2 \):
\[ (-6464)^2 = 41717456 \]
Now calculating \( 4 \cdot 16 \cdot 6060 \):
\[ 4 \cdot 16 = 64 \] \[ 64 \cdot 6060 = 387840 \]
Now, substituting into the discriminant:
\[ b^2 - 4ac = 41717456 - 387840 = 41329616 \]
Now we can compute \( \sqrt{41329616} \):
\[ \sqrt{41329616} = 6428 \]
Now substituting this back into the quadratic formula:
\[ t = \frac{-(-6464) \pm 6428}{2 \cdot 16} \]
This simplifies to:
\[ t = \frac{6464 \pm 6428}{32} \]
Calculating the two possible values for \( t \):
- \( t = \frac{6464 + 6428}{32} = \frac{12892}{32} = 403.5 \)
- \( t = \frac{6464 - 6428}{32} = \frac{36}{32} = 1.125 \)
Thus, the ball reaches the height of 7272 feet at:
- \( t = 1.125 \) seconds, and
- \( t = 403.5 \) seconds.
Therefore, it takes 1.125 seconds for the ball to first reach the height of 7272 feet.
So the final answer is:
It takes 1.125 seconds for the ball to reach the height 7272 feet (and also 403.5 seconds later on).