To find out how long after the ball is thrown it reaches a height of 112112 feet, we start with the height equation:
\[ S(t) = 44 + 9696t - 16t^2 \]
We set \( S(t) \) equal to 112112:
\[ 112112 = 44 + 9696t - 16t^2 \]
Now, rearranging the equation gives us:
\[ 0 = -16t^2 + 9696t + 44 - 112112 \]
Simplifying further:
\[ 0 = -16t^2 + 9696t - 112068 \]
To make calculations easier, we can multiply the equation by -1:
\[ 0 = 16t^2 - 9696t + 112068 \]
Now we can solve this quadratic equation \( 16t^2 - 9696t + 112068 = 0 \) using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where:
- \( a = 16 \)
- \( b = -9696 \)
- \( c = 112068 \)
Calculating the discriminant (\( b^2 - 4ac \)):
\[ b^2 = (-9696)^2 = 93931376 \] \[ 4ac = 4 \times 16 \times 112068 = 7161088 \] \[ b^2 - 4ac = 93931376 - 7161088 = 86770288 \]
Now we compute:
\[ t = \frac{9696 \pm \sqrt{86770288}}{32} \]
Calculating \( \sqrt{86770288} \):
\[ \sqrt{86770288} \approx 2944.42 \]
Now substituting back into the formula:
\[ t = \frac{9696 \pm 2944.42}{32} \]
Calculating the two possible values for \( t \):
-
Using the positive root: \[ t_1 = \frac{9696 + 2944.42}{32} \approx \frac{12640.42}{32} \approx 394.38 \]
-
Using the negative root: \[ t_2 = \frac{9696 - 2944.42}{32} \approx \frac{6751.58}{32} \approx 210.36 \]
Thus, the time it takes for the ball to reach the height of 112112 feet is approximately:
\[ t \approx 394.38, 210.36 \]
For completeness, depending on the context, it is usually only the positive time that is considered meaningful in these problems.
Therefore, the answer for \( t \) is approximately:
394.38 seconds and 210.36 seconds.
So you would need to state:
It takes approximately 394.38 seconds (or 210.36 seconds) for the ball to reach the height of 112112 feet.
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